Elementary Differential Geometry · 2019-03-02 · Elementary Differential Geometry R evised Second Edition Barrett OÕNeill Department of Mathematics University of California, Los

Elementary Differential Geometry

Revised Second Edition



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Barrett O’NeillDepartment of Mathematics

University of California, Los Angeles

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Library of Congress Cataloging-in-Publication DataO’Neill, Barrett.

Elementary differential geometry / Barrett O’Neill.—Rev. 2nd ed.p. cm.

Includes bibliographical references and index.ISBN-13: *978-0-12-088735-4 (acid-free paper)ISBN-10: 0-12-088735-5 (acid-free paper)1. Geometry, Differential. I. Title.

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Contents

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Preface to the Revised Second Edition ix

Introduction 1

1. Calculus on Euclidean Space1.1. Euclidean Space 31.2. Tangent Vectors 61.3. Directional Derivatives 111.4. Curves in R3 161.5. 1-Forms 231.6. Differential Forms 281.7. Mappings 341.8. Summary 41

2. Frame Fields2.1. Dot Product 432.2. Curves 522.3. The Frenet Formulas 582.4. Arbitrary-Speed Curves 692.5. Covariant Derivatives 812.6. Frame Fields 842.7. Connection Forms 882.8. The Structural Equations 942.9. Summary 99

v

3. Euclidean Geometry3.1. Isometries of R3 1003.2. The Tangent Map of an Isometry 1073.3. Orientation 1103.4. Euclidean Geometry 1163.5. Congruence of Curves 1213.6. Summary 128

4. Calculus on a Surface4.1. Surfaces in R3 1304.2. Patch Computations 1394.3. Differentiable Functions and Tangent Vectors 1494.4. Differential Forms on a Surface 1584.5. Mappings of Surfaces 1664.6. Integration of Forms 1744.7. Topological Properties of Surfaces 1844.8. Manifolds 1934.9. Summary 201

5. Shape Operators5.1. The Shape Operator of M Ã R3 2025.2. Normal Curvature 2095.3. Gaussian Curvature 2165.4. Computational Techniques 2245.5. The Implicit Case 2355.6. Special Curves in a Surface 2405.7. Surfaces of Revolution 2525.8. Summary 262

6. Geometry of Surfaces in R3

6.1. The Fundamental Equations 2636.2. Form Computations 2696.3. Some Global Theorems 2736.4. Isometries and Local Isometries 2816.5. Intrinsic Geometry of Surfaces in R3 2896.6. Orthogonal Coordinates 2946.7. Integration and Orientation 2976.8. Total Curvature 3046.9. Congruence of Surfaces 3146.10. Summary 319

vi Contents

7. Riemannian Geometry7.1. Geometric Surfaces 3217.2. Gaussian Curvature 3297.3. Covariant Derivative 3377.4. Geodesics 3467.5. Clairaut Parametrizations 3537.6. The Gauss-Bonnet Theorem 3647.7. Applications of Gauss-Bonnet 3767.8. Summary 386

8. Global Structure of Surfaces8.1. Length-Minimizing Properties of Geodesics 3888.2. Complete Surfaces 4008.3. Curvature and Conjugate Points 4058.4. Covering Surfaces 4168.5. Mappings That Preserve Inner Products 4258.6. Surfaces of Constant Curvature 4338.7. Theorems of Bonnet and Hadamard 4428.8. Summary 449

Appendix: Computer Formulas 451

Bibliography 467

Answers to Odd-Numbered Exercises 468

Index 495

Contents vii

Preface to the Revised Second Edition

ix

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This book is an elementary account of the geometry of curves and surfaces.It is written for students who have completed standard courses in calculusand linear algebra, and its aim is to introduce some of the main ideas of dif-ferential geometry.

The language of the book is established in Chapter 1 by a review of thecore content of differential calculus, emphasizing linearity. Chapter 2describes the method of moving frames, which is introduced, as in elemen-tary calculus, to study curves in space. (This method turns out to apply withequal efficiency to surfaces.) Chapter 3 investigates the rigid motions of space,in terms of which congruence of curves and surfaces is defined in the sameway as congruence of triangles in the plane.

Chapter 4 requires special comment. One weakness of classical differentialgeometry is its lack of any adequate definition of surface. In this chapter wedecide just what a surface is, and show that every surface has a differentialand integral calculus of its own, strictly analogous to the familiar calculus ofthe plane. This exposition provides an introduction to the notion of differ-entiable manifold, which is the foundation for those branches of mathemat-ics and its applications that are based on the calculus.

The next two chapters are devoted to the geometry of surfaces in 3-space.Chapter 5 measures the shape of a surface and derives basic geometric invari-ants, notably Gaussian curvature. Intuitive and computational aspects arestressed to give geometrical meaning to the theory in Chapter 6.

In the final two chapters, although our methods are unchanged, there is aradical shift of viewpoint. Roughly speaking, we study the geometry of asurface as seen by its inhabitants, with no assumption that the surface can befound in ordinary three-dimensional space. Chapter 7 is dominated by cur-vature and culminates in the Gauss-Bonnet theorem and its geometric andtopological consequences. In particular, we use the Gauss-Bonnet theorem to

prove the Poincaré-Hopf theorem, which relates the singularities of a vectorfield on M to the topology of M.

Chapter 8 studies the local and global properties of geodesics. Full devel-opment of the global properties requires the notion of covering surface. Withit, we can give a comprehensive survey of the surfaces of constant Gaussiancurvature and prove the theorems of Bonnet and Hadamard on, respectively,positive and nonnegative curvature.

No branch of mathematics makes a more direct appeal to the intuitionthan geometry. I have sought to emphasize this by a large number of illus-trations that form an integral part of the text.

Each chapter of the book is divided into sections, and in each section asingle sequence of numbers designates collectively the theorems, lemmas,examples, and so on. Each section ends with a set of exercises; these rangefrom routine checks of comprehension to moderately challenging problems.

In this revision, the structure of the text, including the numbering of itscontents, remains the same, but there are many changes around this frame-work. The most significant are, first, correction of all known errors; second,a better way of referencing exercises (the most common reference); third,general improvement of the exercises. These improvements include deletionof a few unreasonably difficult exercises, simplification of others, and fulleranswers to odd-numbered ones.

In teaching from earlier versions of this book, I have usually covered thebackground material in Chapter 1 rather rapidly and not devoted any class-room time to Chapter 3. A short course in the geometry of curves and sur-faces in 3-space might consist of Chapter 2 (omit Sec. 8), Chapter 4 (omitSec. 8), Chapter 5, Chapter 6 (covering Secs. 6–9 lightly), and a leap to Section6 of Chapter 7: the Gauss-Bonnet theorem. This is essentially the content ofa traditional undergraduate course in differential geometry, with clarificationof the notions of surface and mapping.

Such a course, however, neglects the shift of viewpoint mentioned earlier,in which the geometric concept of surface evolved from a shape in 3-space toan independent entity—a two-dimensional Riemannian manifold.

This development is important from a practical viewpoint since it makessurface theory applicable throughout the range of scientific applicationswhere 2-parameter objects appear that meet the requisite conditions—forexample, in the four-dimensional manifolds of general relativity.

Such a surface is logically simpler than a surface in 3-space since it is con-structed (at the start of Chapter 7) by discarding effects of Euclidean space.However, readers can neglect this transition and—as suggested for the Gauss-Bonnet theorem—proceed directly to most of the topics considered in thefinal two chapters, for example, properties of geodesics (length-minimization

x Preface to the Revised Second Edition

and completeness), singularities of vector fields, and the theorems of Bonnetand Hadamard.

For readers with access to a computer containing either the Mathematicaor Maple computation system, I have included some forty computer exer-cises. These offer an opportunity to amplify the text in various ways.

Previous computer experience is not required. The Appendix contains asummary of the syntaxes of the most recent versions of Mathematica andMaple, together with a list of explicit computer commands covering the basicgeometry of curves and surfaces. Further commands appear in the answersto exercises.

It is important to go, step by step, through the hand calculation of theGaussian curvature of a parametrized surface, but once this is understood,repetition becomes tedious. A surface in R3 given only by a formula is seldomeasy to sketch. But using computer commands, a picture of a surface can bedrawn and its curvature computed, often in no more than a few seconds.Analogous remarks hold for space curves.

Among other applications appearing in the exercises, the most valuable,since unreachable for humans, is the numerical solution of differential equa-tions—and the plotting of these solutions.

This book would not have been possible without generous contributionsby Allen B. Altman and Joseph E. Borzellino.

Barrett O’Neill

Preface to the Revised Second Edition xi



Introduction

1

This book presupposes a reasonable knowledge of elementary calculus andlinear algebra. It is a working knowledge of the fundamentals that is actu-ally required. The reader will, for example, frequently be called upon to usethe chain rule for differentiation, but its proof need not concern us.

Calculus deals mostly with real-valued functions of one or more variables,linear algebra with functions (linear transformations) from one vector spaceto another. We shall need functions of these and other types, so we give heregeneral definitions that cover all types.

A set S is a collection of objects that are called the elements of S. A set Ais a subset of S provided each element of A is also an element of S.

A function f from a set D to a set R is a rule that assigns to each elementx of D a unique element f(x) of R. The element f (x) is called the value of fat x. The set D is called the domain of f ; the set R is sometimes called therange of f. If we wish to emphasize the domain and range of a function f,the notation f : D Æ R is used. Note that the function is denoted by a singleletter, say f, while f(x) is merely a value of f.

Many different terms are used for functions—mappings, transformations,correspondences, operators, and so on. A function can be described invarious ways, the simplest case being an explicit formula such as

which we may also write as x Æ 3x2 + 1.If both f1 and f2 are functions from D to R, then f1 = f2 means that

f1(x) = f2(x) for all x in D. This is not a definition, but a logical consequenceof the definition of function.

Let f : D Æ R and g: E Æ S be functions. In general, the image of f is the subset of R consisting of all elements of the form f(x); it is usuallydenoted by f(D). If this image happens to be a subset of the domain E of g,

f x x( ) = +3 12 ,

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it is possible to combine these two functions to get the composite functiong( f ): D Æ S. By definition, g( f ) is the function whose value at each elementx of D is the element g( f (x)) of S.

If f: D Æ R is a function and A is a subset of D, then the restriction of fto A is the function f |A: A Æ R defined by the same rule as f, but appliedonly to elements of A. This seems a rather minor change, but the functionf |A may have properties quite different from f itself.

Here are two vital properties that a function may possess. A function f : D Æ R is one-to-one provided that if x and y are any elements of D suchthat x π y, then f(x) π f(y). A function f: D Æ R is onto (or carries D ontoR) provided that for every element y of R there is at least one element x ofD such that f(x) = y. In short, the image of f is the entire set R. For example,consider the following functions, each of which has the real numbers as bothdomain and range:

(1) The function x Æ x3 is both one-to-one and onto.(2) The exponential function x Æ ex is one-to-one, but not onto.(3) The function x Æ x3 + x2 is onto, but not one-to-one.(4) The sine function x Æ sin x is neither one-to-one nor onto.

If a function f : D Æ R is both one-to-one and onto, then for each elementy of R there is one and only one element x such that f(x) = y. By defining f -1(y) = x for all x and y so related, we obtain a function f -1: R Æ D calledthe inverse of f. Note that the function f -1 is also one-to-one and onto, andthat its inverse function is the original function f.

Here is a short list of the main notations used throughout the book, inorder of their appearance in Chapter 1:

p, q . . . . . . . . . . . . . . . . . . . . . points (Section 1.1)f, g . . . . . . . . . . . . . . . . . . . . . real-valued functions (Section 1.1)v, w . . . . . . . . . . . . . . . . . . . . . tangent vectors (Section 1.2)V, W . . . . . . . . . . . . . . . . . . . . vector fields (Section 1.2)a, b . . . . . . . . . . . . . . . . . . . . . curves (Section 1.4)f, y . . . . . . . . . . . . . . . . . . . . . differential forms (Section 1.5)F, G . . . . . . . . . . . . . . . . . . . . . mappings (Section 1.7)

In Chapter 1 we define these concepts for Euclidean 3-space. (Extension toarbitrary dimensions is virtually automatic.) In Chapter 4 we show how theseconcepts can be adapted to a surface.

A few references are given to the brief bibliography at the end of the book;these are indicated by initials in square brackets.

2 Introduction

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▲Chapter 1

Calculus on Euclidean Space

3

As mentioned in the Preface, the purpose of this initial chapter is to estab-lish the mathematical language used throughout the book. Much of what wedo is simply a review of that part of elementary calculus dealing with differ-entiation of functions of three variables and with curves in space. Our defi-nitions have been formulated so that they will apply smoothly to the laterstudy of surfaces.

1.1 Euclidean Space

Three-dimensional space is often used in mathematics without being formallydefined. Looking at the corner of a room, one can picture the familiar processby which rectangular coordinate axes are introduced and three numbers aremeasured to describe the position of each point. A precise definition thatrealizes this intuitive picture may be obtained by this device: instead of sayingthat three numbers describe the position of a point, we define them to be apoint.

1.1 Definition Euclidean 3-space R3 is the set of all ordered triples of realnumbers. Such a triple p = ( p1, p2, p3) is called a point of R3.

In linear algebra, it is shown that R3 is, in a natural way, a vector spaceover the real numbers. In fact, if p = ( p1, p2, p3) and q = (q1, q2, q3) are pointsof R3, their sum is the point

p q+ = + + +( )p q p q p q1 1 2 2 3 3, , .

The scalar multiple of a point p = ( p1, p2, p3) by a number a is the point

It is easy to check that these two operations satisfy the axioms for a vectorspace. The point 0 = (0, 0, 0) is called the origin of R3.

Differential calculus deals with another aspect of R3 starting with thenotion of differentiable real-valued functions on R3. We recall some fundamentals.

1.2 Definition Let x, y, and z be the real-valued functions on R3 suchthat for each point p = ( p1, p2, p3)

These functions x, y, z are called the natural coordinate functions of R3. Weshall also use index notation for these functions, writing

Thus the value of the function xi on a point p is the number pi, and so wehave the identity p = ( p1, p2, p3) = (x1(p), x2(p), x3(p)) for each point p of R3.Elementary calculus does not always make a sharp distinction between thenumbers p1, p2, p3 and the functions x1, x2, x3. Indeed the analogous distinc-tion on the real line may seem pedantic, but for higher-dimensional spacessuch as R3, its absence leads to serious ambiguities. (Essentially the same dis-tinction is being made when we denote a function on R3 by a single letter f,reserving f (p) for its value at the point p.)

We assume that the reader is familiar with partial differentiation and itsbasic properties, in particular the chain rule for differentiation of a compos-ite function. We shall work mostly with first-order partial derivatives ∂f /∂x,∂f /∂y, ∂f /∂z and second-order partial derivatives ∂2f /∂x2, ∂2f /∂x∂y, . . . In afew situations, third- and even fourth-order derivatives may occur, but toavoid worrying about exactly how many derivatives we can take in any givencontext, we establish the following definition.

1.3 Definition A real-valued function f on R3 is differentiable (or infi-nitely differentiable, or smooth, or of class C •) provided all partial derivativesof f, of all orders, exist and are continuous.

Differentiable real-valued functions f and g may be added and multipliedin a familiar way to yield functions that are again differentiable and real-

x x x y x z1 2 3= = =, , .

x p y p z pp p p( ) = ( ) = ( ) =1 2 3, , .

a ap ap app = ( )1 2 3, , .

4 1. Calculus on Euclidean Space

valued. We simply add and multiply their values at each point—the formu-las read

The phrase “differentiable real-valued function” is unpleasantly long. Hencewe make the convention that unless the context indicates otherwise, “func-tion” shall mean “real-valued function,” and (unless the issue is explicitlyraised) the functions we deal with will be assumed to be differentiable. We donot intend to overwork this convention; for the sake of emphasis the words“differentiable” and “real-valued” will still appear fairly frequently.

Differentiation is always a local operation: To compute the value of thefunction ∂f/∂x at a point p of R3, it is sufficient to know the values of f at allpoints q of R3 that are sufficiently near p. Thus, Definition 1.3 is undulyrestrictive; the domain of f need not be the whole of R3, but need only be anopen set of R3. By an open set O of R3 we mean a subset of R3 such that if apoint p is in O, then so is every other point of R3 that is sufficiently near p.(A more precise definition is given in Chapter 2.) For example, the set of allpoints p = ( p1, p2, p3) in R3 such that p1 > 0 is an open set, and the function yz logx defined on this set is certainly differentiable, even though its domainis not the whole of R3. Generally speaking, the results in this chapter remainvalid if R3 is replaced by an arbitrary open set O of R3.

We are dealing with three-dimensional Euclidean space only because this isthe dimension we use most often in later work. It would be just as easy towork with Euclidean n-space Rn, for which the points are n-tuples p = ( p1,. . . , pn) and which has n natural coordinate functions x1, . . . , xn. All theresults in this chapter are valid for Euclidean spaces of arbitrary dimensions,although we shall rarely take advantage of this except in the case ofthe Euclidean plane R2. In particular, the results are valid for the real lineR1 = R. Many of the concepts introduced are designed to deal with higherdimensions, however, and are thus apt to be overelaborate when reduced todimension 1.

Exercises

1. Let f = x2y and g = y sinz be functions on R3. Express the followingfunctions in terms of x, y, z:

(a) fg2. (b)

(c) (d)∂∂y

fsin .( )∂∂ ∂

2 fgy z( )

.

∂∂

∂∂

fx

ggy

f+ .

f g f g fg f g+( )( ) = ( ) + ( ) ( )( ) = ( ) ( )p p p p p p, .

1.1 Euclidean Space 5

2. Find the value of the function f = x2y - y2z at each point:(a) (1, 1, 1). (b) (3, -1, ).(c) (a, 1, 1 - a). (d) (t, t2, t3).

3. Express ∂f/∂x in terms of x, y, and z if(a) f = x sin (xy) + ycos (xz).(b) f = sin g, g = eh, h = x2 + y2 + z2.

4. If g1, g2, g3, and h are real-valued functions on R3, then

is the function such that

Express ∂f /∂x in terms of x, y, and z, if h = x2 - yz and(a) f = h(x + y, y2, x + z). (b) f = h(ez, ex+y, ex).(c) f = h(x, -x, x).

1.2 Tangent Vectors

Intuitively, a vector in R3 is an oriented line segment, or “arrow.” Vectors areused widely in physics and engineering to describe forces, velocities, angularmomenta, and many other concepts. To obtain a definition that is both prac-tical and precise, we shall describe an “arrow” in R3 by giving its startingpoint p and the change, or vector v, necessary to reach its end point p + v.Strictly speaking, v is just a point of R3.

2.1 Definition‡ A tangent vector vp to R3 consists of two points of R3: itsvector part v and its point of application p.

We shall always picture vp as the arrow from the point p to the point p + v.For example, if p = (1, 1, 3) and v = (2, 3, 2), then vp runs from (1, 1, 3) to(3, 4, 5) as in Fig. 1.1.

We emphasize that tangent vectors are equal, vp = wq, if and only if theyhave the same vector part, v = w, and the same point of application, p = q.

f h g g gp p p p p( ) = ( ) ( ) ( )( )1 2 3, , for all .†

f h g g g= ( )1 2 3, ,

12


† A consequence is the identity f = f(x, y, z).‡ The term “tangent” in this definition will acquire a more direct geometric meaning in Chapter 4.

Tangent vectors vp and vq with the same vector part, but different points ofapplication, are said to be parallel (Fig. 1.2). It is essential to recognize thatvp and vq are different tangent vectors if p π q. In physics the concept ofmoment of a force shows this clearly enough: The same force v applied atdifferent points p and q of a rigid body can produce quite different rotationaleffects.

2.2 Definition Let p be a point of R3. The set Tp(R3) consisting of alltangent vectors that have p as point of application is called the tangent spaceof R3 at p (Fig. 1.3).

We emphasize that R3 has a different tangent space at each and every oneof its points.

1.2 Tangent Vectors 7

FIG. 1.1

FIG. 1.2 FIG. 1.3

Since all the tangent vectors in a given tangent space have the same pointof application, we can borrow the vector addition and scalar multiplicationof R3 to turn Tp(R3) into a vector space. Explicitly, we define vp + wp to be (v + w)p. and if c is a number we define c(vp) to be (cv)p. This is just the usual“parallelogram law” for addition of vectors, and scalar multiplication by c merely stretches a tangent vector by the factor c—reversing its direction ifc < 0 (Fig. 1.4).

These operations on the tangent space Tp(R3) make it a vector space iso-morphic to R3 itself. Indeed, it follows immediately from the definitions abovethat for a fixed point p, the function v Æ vp is a linear isomorphism from R3

to Tp(R3)—that is, a linear transformation that is one-to-one and onto.A standard concept in physics and engineering is that of a force field. The

gravitational force field of the earth, for example, assigns to each point ofspace a force (vector) directed at the center of the earth.

2.3 Definition A vector field V on R3 is a function that assigns to eachpoint p of R3 a tangent vector V (p) to R3 at p.

Roughly speaking, a vector field is just a big collection of arrows, one ateach point of R3.

There is a natural algebra of vector fields. To describe it, we first reexam-ine the familiar notion of addition of real-valued functions f and g. It is pos-sible to add f and g because it is possible to add their values at each point.The same is true of vector fields V and W. At each point p, the values V(p)and W(p) are in the same vector space—the tangent space Tp(R3)—hence wecan add V(p) and W(p). Consequently, we can add V and W by adding their


FIG. 1.4

values at each point. The formula for this addition is thus the same as foraddition of functions,

This scheme occurs over and over again. We shall call it the pointwise princi-ple: If a certain operation can be performed on the values of two functionsat each point, then that operation can be extended to the functions them-selves; simply apply it to their values at each point.

For example, we invoke the pointwise principle to extend the operation ofscalar multiplication (on the tangent spaces of R3). If f is a real-valued func-tion on R3 and V is a vector field on R3, then f V is defined to be the vectorfield on R3 such that

Our aim now is to determine in a concrete way just what vector fields looklike. For this purpose we introduce three special vector fields that will serveas a “basis” for all vector fields.

2.4 Definition Let U1, U2, and U3 be the vector fields on R3 such that

for each point p of R3 (Fig. 1.5). We call U1, U2, U3—collectively—the naturalframe field on R3.

Thus, Ui (i = 1, 2, 3) is the unit vector field in the positive xi direction.

2.5 Lemma If V is a vector field on R3, there are three uniquely deter-mined real-valued functions, v1, v2, v3 on R3 such that

U

U

U

p

p

p

1

2

3

1 0 0

0 1 0

0 0 1

p

p

p

( ) = ( )( ) = ( )( ) = ( )

, ,

, ,

, ,

fV f V( )( ) = ( ) ( )p p p p for all .

V W V W+( )( ) = ( ) + ( )p p p .

1.2 Tangent Vectors 9

FIG. 1.5

The functions v1, v2, v3 are called the Euclidean coordinate functions of V.

Proof. By definition, the vector field V assigns to each point p a tangentvector V(p) at p. Thus, the vector part of V(p) depends on p, so we writeit (v1(p), v2(p), v3(p)). (This defines v1, v2, and v3 as real-valued functions onR3.) Hence

for each point p (Fig. 1.6). By our (pointwise principle) definitions, thismeans that the vector fields V and viUi have the same (tangent vector)value at each point. Hence V = viUi. �

This last sentence uses two of our standard conventions: viUi means sumover i = 1, 2, 3; the symbol (�) indicates the end of a proof.

The tangent-vector identity (a1, a2, a3)p = aiUi(p) appearing in this proofwill be used very often.

Computations involving vector fields may always be expressed in terms oftheir Euclidean coordinate functions. For example, addition and multiplica-tion by a function, are expressed in terms of coordinates by

vU wU v w U

f vU fv U

i i i i i i i

i i i i

Â Â ÂÂ Â

+ = +( )

( ) = ( ),

.

Â

ÂÂ

Â

V v v v

v v v

v U v U v U

p

p p p

p p p p

p p p

p p p p p p

( ) = ( ) ( ) ( )( )= ( )( ) + ( )( ) + ( )( )= ( ) ( ) + ( ) ( ) + ( ) ( )

1 2 3

1 2 3

1 1 2 2 3 3

1 0 0 0 1 0 0 0 1

, ,

, , , , , ,

V vU vU vU= + +1 1 2 2 3 3.


FIG. 1.6

Since this is differential calculus, we shall naturally require that the variousobjects we deal with be differentiable. A vector field V is differentiable pro-vided its Euclidean coordinate functions are differentiable (in the sense ofDefinition 1.3). From now on, we shall understand “vector field” to mean“differentiable vector field.”

Exercises

1. Let v = (-2, 1, -1) and w = (0, 1, 3).(a) At an arbitrary point p, express the tangent vector 3vp - 2wp as a linearcombination of U1(p), U2(p), U3(p).(b) For p = (1, 1, 0), make an accurate sketch showing the four tangentvectors vp, wp, -2vp, and vp + wp.

2. Let V = xU1 + yU2 and W = 2x2U2 - U3. Compute the vector field W - xV, and find its value at the point p = (-1, 0, 2).

3. In each case, express the given vector field V in the standard form viUi.(a) 2z2U1 = 7V + xyU3.(b) V(p) = ( p1, p3 - p1, 0)p for all p.(c) V = 2(xU1 + yU2) - x(U1 - y2U3).(d) At each point p, V(p) is the vector from the point ( p1, p2, p3) to thepoint (1 + p1, p2p3, p2).(e) At each point p, V(p) is the vector from p to the origin.

4. If V = y2U1 - x2U3 and W = x2U1 - zU2, find functions f and g suchthat the vector field f V + gW can be expressed in terms of U2 and U3 only.

5. Let V1 = U1 - xU3, V2 = U2, and V3 = xU1 + U3.(a) Prove that the vectors V1(p), V2(p), V3(p) are linearly independent ateach point of R3.(b) Express the vector field xU1 + yU2 + zU3 as a linear combination ofV1, V2, V3.

1.3 Directional Derivatives

Associated with each tangent vector vp to R3 is the straight line t Æ p + tv(see Example 4.2). If f is a differentiable function on R3, then t Æ f(p + tv)is an ordinary differentiable function on the real line. Evidently the deriva-tive of this function at t = 0 tells the initial rate of change of f as p movesin the v direction

Â

1.3 Directional Derivatives 11

3.1 Definition Let f be a differentiable real-valued function on R3, andlet vp be a tangent vector to R3. Then the number

is called the derivative of f with respect to vp.This definition appears in elementary calculus with the additional restric-

tion that vp be a unit vector. Even though we do not impose this restriction,we shall nevertheless refer to vp[ f ] as a directional derivative.

For example, we compute vp[ f ] for the function f = x2yz, with p = (1, 1, 0)and v = (1, 0, -3). Then

describes the line through p in the v direction. Evaluating f along this line,we get

Now,

hence at t = 0, we find vp[ f ] = -3. Thus, in particular, the function f is initially decreasing as p moves in the v direction.

The following lemma shows how to compute vp[ f ] in general, in terms ofthe partial derivatives of f at the point p.

3.2 Lemma If vp = (v1, v2, v3)p is a tangent vector to R3, then

Proof. Let p = ( p1, p2, p3); then

We use the chain rule to compute the derivative at t = 0 of the function

Since

ddt

p tv vi i i+( ) = ,

f t f p tv p tv p tvp v+( ) = + + +( )1 1 2 2 3 3, , .

p v+ = + + +( )t p tv p tv p tv1 1 2 2 3 3, , .

v pp ii

f vf

x[ ] =

∂∂

( )Â .

ddt

f t t tp v+( )( ) = - - -3 12 9 2;

f t t t t t tp v+( ) = +( ) ◊ ◊ -( ) = - - -1 1 3 3 6 32 2 3.

p v+ = ( ) + -( ) = + -( )t t t t1 1 0 1 0 3 1 1 3, , , , , ,

v p vp tfddt

f t[ ] = +( )( ) =0


we obtain

�

Using this lemma, we recompute vp[ f ] for the example above. Since f = x2yz, we have

Thus, at the point p = (1, 1, 0),

Then by the lemma,

as before.The main properties of this notion of derivative are as follows.

3.3 Theorem Let f and g be functions on R3, vp and wp tangent vectors,a and b numbers. Then

(1) (avp + bwp)[ f ] = avp[ f ] + bwp[ f ].(2) vp[af + bg] = avp[ f ] + bvp[g].(3) vp[ fg] = vp[ f ] .g(p) + f(p) .vp[g].

Proof. All three properties may be deduced easily from the precedinglemma. For example, we prove (3). By the lemma, if v = (v1, v2, v3), then

But

Hence

�

v p p p p

p p p p

v p p v

p ii i

ii

ii

p p

fg vf

xg f

gx

vf

xg f v

gx

f g f g

[ ] =∂∂

( ) ◊ ( ) + ( ) ◊∂∂

( )ÊËÁ

ˆ¯

=∂∂

( )ÊËÁ

ˆ¯

( ) + ( ) ∂∂

( )ÊËÁ

ˆ¯

= [ ] ◊ ( ) + ( ) ◊ [ ]

Â

Â Â

.

∂( )∂

=∂∂

◊ + ◊∂∂

fgx

fx

g fgxi i i

.

v pp ii

fg vfgx

[ ] =∂( )∂

( )Â .

v p f[ ] = + + -( ) = -0 0 3 1 3,

∂∂

( ) =∂∂

( ) =∂∂

( ) =fx

fy

fz

p p p0 0 1, , and .

∂∂

=∂∂

=∂∂

=fx

xyzfy

x zfz

x y2 2 2, , .

v p v pp ti

ifddt

f tfx

v[ ] = +( )( ) =∂∂

( )= Â| .0


The first two properties in the preceding theorem may be summarized bysaying that vp[ f ] is linear in vp and in f. The third property, as its proof makesclear, is essentially just the usual Leibniz rule for differentiation of a product.No matter what form differentiation may take, it will always have suitable linearand Leibnizian properties.

We now use the pointwise principle to define the operation of a vector fieldV on a function f. The result is the real-valued function V[ f ] whose value ateach point p is the number V(p)[ f ], that is, the derivative of f with respect tothe tangent vector V(p) at p. This process should be no surprise, since for afunction f on the real line, one begins by defining the derivative of f at apoint—then the derivative function df/dx is the function whose value at eachpoint is the derivative at that point. Evidently, the definition of V[ f ] is strictlyanalogous. In particular, if U1, U2, U3 is the natural frame field on R3, thenUi [ f ] = ∂f/∂xi. This is an immediate consequence of Lemma 3.2. Forexample, U1(p) = (1, 0, 0)p; hence

which is precisely the definition of (∂f/∂x1)(p). This is true for all points p = (p1, p2, p3); hence U1[ f ] = ∂f/∂x1.

We shall use this notion of directional derivative more in the case of vectorfields than for individual tangent vectors.

3.4 Corollary If V and W are vector fields on R3 and f, g, h are real-valued functions, then

(1) ( fV + gW)[h] = fV [h] + gW [h].(2) V [af + bg] = aV [ f ] + bV [g], for all real numbers a and b.(3) V [ fg] = V[ f ] .g + f .V [g].

Proof. The pointwise principle guarantees that to derive these propertiesfrom Theorem 3.3 we need only be careful about the placement of paren-theses. For example, we prove the third formula. By definition, the valueof the function V[ fg] at p is V (p)[ fg]. But by Theorem 3.3 this is

�

If the use of parentheses here seems extravagant, we remind the reader thata meticulous proof of Leibniz’s formula

V f g f V g V f g f V g

V f g f V g

p p p p p p p p

p

( )[ ]◊ ( ) + ( )◊ ( )[ ] = [ ]( )◊ ( ) + ( )◊ [ ]( )= [ ]◊ + ◊ [ ]( )( ).

U fddt

f p t p p t1 1 2 3 0p( )[ ] = +( )( ) =, , ,


must involve the same shifting of parentheses.Note that the linearity of V[ f ] in V and f is for functions as “scalars” in

the first formula in Corollary 3.4 but only for numbers as “scalars” in thesecond. This stems from the fact that fV signifies merely multiplication, butV[ f ] is differentiation.

The identity Ui [ f ] = ∂f/∂xi makes it a simple matter to carry out explicitcomputations. For example, if V = xU1 - y2U3 and f = x2y + z3, then

3.5 Remark Since the subscript notation vp for a tangent vector is some-what cumbersome, from now on we shall frequently omit the point of appli-cation p from the notation. This can cause no confusion, since v and w willalways denote tangent vectors, and p and q points of R3. In many situations(for example, Definition 3.1) the point of application is crucial, and will beindicated by using either the old notation vp or the phrase “a tangent vectorv to R3 at p.”

Exercises

1. Let vp be the tangent vector to R3 with v = (2, -1, 3) and p = (2, 0, -1).Working directly from the definition, compute the directional derivative vp[ f ],where

(a) f = y2z. (b) f = x7.(c) f = ex cos y.

2. Compute the derivatives in Exercise 1 using Lemma 3.2.

3. Let V = y2U1 - xU3, and let f = xy, g = z3. Compute the functions(a) V[ f ]. (b) V[g].(c) V[ fg]. (d) fV[g] - gV[ f ].(e) V[ f 2 + g2]. (f) V[V[ f ]].

4. Prove the identity V = V [xi ]Ui, where x1, x2, x3 are the natural coor-dinate functions. (Hint: Evaluate V = viUi on xj.)

5. If V [ f ] = W[ f ] for every function f on R3, prove that V = W.

ÂÂ

V f xU x y xU z y U x y y U z

x xy y z x y y z

[ ] = [ ] + [ ] - [ ] - [ ]= ( ) + - - ( ) = -

12

13 2

32 2

33

2 2 2 2 22 0 0 3 2 3 .

ddx

fgdfdx

g fdgdx

( ) = ◊ + ◊


1.4 Curves in R3

Let I be an open interval in the real line R. We shall interpret this liberallyto include not only the usual finite open interval a < t < b (a, b real numbers),but also the infinite types a < t (a half-line to +•), t < b (a half-line to -•),and also the whole real line.

One can picture a curve in R3 as a trip taken by a moving point a. At each“time” t in some open interval, a is located at the point

in R3. In rigorous terms then, a is a function from I to R3, and the real-valuedfunctions a1, a2, a3 are its Euclidean coordinate functions. Thus we write a = (a1, a, a3), meaning, of course, that

We define the function a to be differentiable provided its (real-valued) co-ordinate functions are differentiable in the usual sense.

4.1 Definition A curve in R3 is a differentiable function a : I Æ R3 froman open interval I into R3.

We shall give several examples of curves, which will be used in Chapter 2to experiment with results on the geometry of curves.

4.2 Example (1) Straight line. A line is the simplest type of curve inEuclidean space; its coordinate functions are linear (in the sense t Æ at + b,not in the homogeneous sense t Æ at). Explicitly, the curve a: R Æ R3 suchthat

is the straight line through the point p = a(0) in the q direction.

(2) Helix. (Fig. 1.7). The curve t Æ (acos t,a sin t,0) travels around a circleof radius a > 0 in the xy plane of R3. If we allow this curve to rise (or fall)at a constant rate, we obtain a helix a: R Æ R3, given by the formula

where a > 0, b π 0.

a t a t a t bt( ) = ( )cos , sin ,

a t t p tq p tq p tq( ) = + = + + +( ) π( )p q q1 1 2 2 3 3 0, ,

a a a at t t t t I( ) = ( ) ( ) ( )( )1 2 3, ., for all in

a a a at t t t( ) = ( ) ( ) ( )( )1 2 3, ,


(3) The curve

has a noteworthy property: Let C be the cylinder in R3 over the circle in thexy plane with center at (1, 0, 0) and radius 1. Then a perpetually travels theroute sliced from C by the sphere with radius 2 and center at the origin.A segment of this route is shown in Fig. 1.8.

(4) The curve a: R Æ R3 such that

shares with the helix in (2) the property of rising constantly. However, it liesover the hyperbola xy = 1 in the xy plane instead of over a circle.

(5) The 3-curve a: R Æ R3 is defined by

If the coordinate functions of a curve are simple enough, its shape in R3 canbe found, at least approximately, by plotting a few points. We could get a rea-sonable picture of curve a for 0 � t � 1 by computing a(t) for t = 0, 1/10, 1/2,9/10, 1.

If we visualize a curve a in R3 as a moving point, then at every time t thereis a tangent vector at the point a(t) that gives the instantaneous velocity ofa at that time. ◆

a t t t t t t( ) = - +( )3 3 33 2 3, , .

a t e e tt t( ) = ( )-, , 2

Â

a t t tt

t( ) = +ÊË

ˆ¯1 2

2cos sin sin, , for all

1.4 Curves in R3 17

FIG. 1.7 FIG. 1.8

4.3 Definition Let a: I Æ R3 be a curve in R3 with a = (a1, a2, a3). Foreach number t in I, the velocity vector of a at t is the tangent vector

at the point a(t) in R3 (Fig. 1.9).

This definition can be interpreted geometrically as follows. The derivativeat t of a real-valued function f on R is given by

This formula still makes sense if f is replaced by a curve a = (a1, a2, a3). Infact,

This is the vector from a(t) to a(t + Dt), scalar multiplied by 1/Dt (Fig. 1.10).Now, as Dt gets smaller, a(t + Dt) approaches a(t), and in the limit as

Dt Æ 0, we get a vector tangent to the curve a at the point a(t), namely,

As the figure suggests, the point of application of this vector must be thepoint a(t). Thus the standard limit operation for derivatives gives rise to ourdefinition of the velocity of a curve.

ddt

tddt

tddt

ta a a1 2 3( ) ( ) ( )Ê

Ëˆ¯, , .

1 1 1

2 2 3 3

DD

DD

DD

DD

tt t t

t t tt

t t tt

t t tt

a aa a

a a a a

+( ) - ( )( ) =+( ) - ( )Ê

Ë

+( ) - ( ) +( ) - ( )ˆ¯

,

, .

dfdt

tf t t f t

tt( ) =

+( ) - ( )Æ

lim .D

DD0

¢( ) = ( ) ( ) ( )ÊË

ˆ¯ ( )

aa a a

a

tddt

tddt

tddt

tt

1 2 3, ,


FIG. 1.9

An application of the identity

to the velocity vector a ¢(t) at t yields the alternative formula

For example, the velocity of the straight line a(t) = p + tq is

The fact that a is straight is reflected in the fact that all its velocity vectorsare parallel; only the point of application changes as t changes.

For the helix

the velocity is

The fact that the helix rises constantly is shown by the constancy of the zcoordinate of a ¢(t).

Given any curve, it is easy to construct new curves that follow the sameroute.

4.4 Definition Let a: I Æ R3 be a curve. If h: J Æ I is a differentiablefunction on an open interval J, then the composite function

is a curve called a reparametrization of a by h.

For each s Œ J, the new curve b is at the point b(s) = a(h(s)) reached bya at h(s) in I (Fig. 1.11). Thus b represents a different trip over at least partof the route of a.

b a= ( ) Æh J: R3

¢( ) = -( ) ( )a at a t a t bt

sin cos ., ,

a t a t a t bt( ) = ( )cos sin ,, ,

¢( ) = ( ) =( ) ( )a a at q q qt t1 2 3, , q .

¢( ) = ( ) ( )( )Âaa

atddt

t U tii .

v v v vUp i i1 2 3, ,( ) = ( )Â p

1.4 Curves in R3 19

FIG. 1.10

To compute the coordinates of b, simply substitute t = h(s) into the co-ordinates a1(t), a2(t), a3(t) of a. For example, suppose

If h(s) = s2 on J: 0 < s < 2, then the reparametrized curve is

The following lemma relates the velocities of a curve and of a repara-metrization.

4.5 Lemma If b is the reparametrization of a by h, then

Proof. If a = (a1, a2, a3), then

Using the “prime” notation for derivatives, the chain rule for a composi-tion of real-valued functions f and g reads (g( f ))¢ = g¢( f ) . f ¢. Thus, in thecase at hand,

By the definition of velocity, this yields

�

According to this lemma, to obtain the velocity of a reparametrization of a by h, first reparametrize a ¢ by h, then scalar multiply by the derivativeof h.

Since velocities are tangent vectors, we can take the derivative of a func-tion with respect to a velocity.

¢( ) = ( )¢( )= ¢ ( )( )◊ ¢( ) ¢ ( )( )◊ ¢( ) ¢ ( )( )◊ ¢( )( )= ¢( ) ¢ ( )( )

b a

a a a

a

s h s

h s h s h s h s h s h s

h s h s

1 2 3, ,

.

a ai ih s h s h s( )¢( ) = ¢ ( )( )◊ ¢( ).

b a a a as h s h s h s h s( ) = ( )( ) = ( )( ) ( )( ) ( )( )( 1 2 3, , .

¢( ) = ( )( ) ¢ ( )( )b as dh ds s h s .

b a as h s s s s s( ) = ( )( ) = ( ) = -( )2 3 21, , .

a t t t t t on I t( ) = -( ) < <, , 1 0 4: .


FIG. 1.11

4.6 Lemma Let a be a curve in R3 and let f be a differentiable functionon R3. Then

Proof. Since

we conclude from Lemma 3.2 that

But the composite function f(a) may be written f(a1, a2, a3), and thechain rule then gives exactly the same result for the derivative of f(a). �

By definition, a ¢(t)[ f ] is the rate of change of f along the line through a(t)in the a ¢(t) direction. (If a ¢(t) π 0, this is the tangent line to a at a(t); seeExercise 9.) The lemma shows that this rate of change is the same as that off along the curve a itself.

Since a curve a: I Æ R3 is a function, it makes sense to say that a is one-to-one; that is, a(t) = a(t1) only if t = t1. Another special property of curvesis periodicity: A curve a: R Æ R3 is periodic if there is a number p > 0 suchthat a(t + p) = a(t) for all t—and the smallest such number p is then calledthe period of a.

From the viewpoint of calculus, the most important condition on a curvea is that it be regular, that is, have all velocity vectors different from zero.Such a curve can have no corners or cusps.

The following remarks about curves (offered without proof ) describeanother familiar way to formulate the concept of “curve.” If f is a differen-tiable real-valued function on R2, let

be the set of all points p in R2 such that f(p) = a. Now, if the partial deriv-atives ∂f/∂x and ∂f/∂y are never simultaneously zero at any point of C, thenC consists of one or more separate “components,” which we shall callCurves.† For example, C: x2 + y2 = r2 is the circle of radius r centered at the

C f a: =

¢( )[ ] =∂∂

( )( ) ( )Âa aa

t ff

xt

ddt

ti

i .

¢ = ÊË

ˆ¯a

a a a

a

ddt

ddt

ddt

1 2 3, , ,

¢( )[ ] =( )( ) ( )aa

t fd f

dtt .

1.4 Curves in R3 21

† The capital C distinguishes this notion from a (parametrized) curve a: I Æ R2.

origin of R2, and the hyperbola C: x2 - y2 = r2 splits into two Curves(“branches”) C1 and C2 as shown in Fig. 1.12.

Every Curve C is the route of many regular curves, called parametrizationsof C. For example, the curve

is a well-known periodic parametrization of the circle given above, and for r > 0 the one-to-one curve

parametrizes the branch x > 0 of the hyperbola.

Exercises

1. Compute the velocity vector of the curve in Example 4.2(3) for arbitraryt and for t = 0, t = p/2, t = p, visualizing those on Fig. 1.8.

2. Find the unique curve such that a(0) = (1, 0, 5) and a¢(t) = (t2, t, et).

3. Find the coordinate functions of the curve b = a(h), where a is the curvein Example 4.2(3) and h(s) = cos-1 (s) on J: 0 < s < 1.

4. Reparametrize the curve a in Example 4.2(4) using h(s) = log s on J: s > 0. Check the equation in Lemma 4.5 in this case by calculating eachside separately.

5. Find the equation of the straight line through the points (1, -3, -1) and (6, 2, 1). Does this line meet the line through the points (-1, 1, 0) and (-5, -1, -1)?

6. Deduce from Lemma 4.6 that in the definition of directional derivative(Def. 3.1) the straight line t Æ p + tv can be replaced by any curve a withinitial velocity vp, that is, such that a(0) = p and a ¢(0) = vp.

b t r t r t( ) = ( )cosh sinh,

a t r t r t( ) = ( )cos sin,


FIG. 1.12

7. (Continuation.)(a) Show that the curves with coordinate functions

all have the same initial velocity vp.(b) If f = x2 - y2 + z2, compute vp[ f ] by calculating d( f(a))/dt at t = 0,using each of three curves in (a).

8. Sketch the following Curves in R2, and find parametrizations for each.(a) C: 4x2 + y2 = 1, (b) C: 3x + 4y = 1,(c) C: y = ex.

9. For a fixed t, the tangent line to a regular curve a at the point a(t) is thestraight line u Æ a(t) + ua ¢(t), where we delete the point of application ofa ¢(t). Find the tangent line to the helix a(t) = (2cos t, 2 sin t, t) at the pointsa(0) and a(p/4).

1.5 1-Forms

If f is a real-valued function on R3, then in elementary calculus the differen-tial of f is usually defined as

It is not always made clear exactly what this formal expression means. In thissection we give a rigorous treatment using the notion of 1-form, and formstend to appear at crucial moments in later work.

5.1 Definition A 1-form f on R3 is a real-valued function on the set ofall tangent vectors to R3 such that f is linear at each point, that is,

for any numbers a, b and tangent vectors v, w at the same point of R3.

We emphasize that for every tangent vector v, a 1-form f defines a realnumber f(v); and for each point p in R3, the resulting function fp: Tp(R3) Æ Ris linear. Thus at each point p, fp is an element of the dual space of Tp(R3). Inthis sense the notion of 1-form is dual to that of vector field.

The sum of 1-forms f and y is defined in the usual pointwise fashion:

f f fa b a bv w v w+( ) = ( ) + ( )

dffx

dxfy

dyfz

dz=∂∂

+∂∂

+∂∂

.

t t t t t t t t t, , , , , , , ,1 2+( ) ( ) ( )sin cos sinh cosh

1.5 1-Forms 23

Similarly, if f is a real-valued function on R3 and f is a 1-form, then ff isthe 1-form such that

for all tangent vectors vp.There is also a natural way to evaluate a 1-form f on a vector field V to

obtain a real-valued function f(V): At each point p the value of f(V) is thenumber f(V(p)). Thus a 1-form may also be viewed as a machine that con-verts vector fields into real-valued functions. If f(V) is differentiable when-ever V is, we say that f is differentiable. As with vector fields, we shall alwaysassume that the 1-forms we deal with are differentiable.

A routine check of definitions shows that f(V) is linear in both f and V;that is,

and

where f and g are functions.Using the notion of directional derivative, we now define a most impor-

tant way to convert functions into 1-forms.

5.2 Definition If f is a differentiable real-valued function on R3, the dif-ferential df of f is the 1-form such that

In fact, df is a 1-form, since by definition it is a real-valued function ontangent vectors, and by (1) of Theorem 3.3 it is linear at each point p. Clearly,df knows all rates of change of f in all directions on R3, so it is not surpris-ing that differentials are fundamental to the calculus on R3.

Our task now is to show that these rather abstract definitions lead to famil-iar results when expressed in terms of coordinates.

5.3 Example 1-Forms on R3. (1) The differentials dx1, dx2, dx3 of thenatural coordinate functions. Using Lemma 3.2 we find

dx x vxx

v vi p p i ji

jj ij i

jj

v v p( ) = [ ] =∂∂

( ) = =ÂÂ d ,

df fp p pv v v( ) = [ ] for all tangent vectors .

f g V f V g Vf y f y+( )( ) = ( ) + ( ),

f f ffV gW f V g W+( ) = ( ) + ( )

f fp pf f( )( ) = ( ) ( )v p v

f y f y+( )( ) = ( ) + ( )v v v vfor all tangent vectors .


where dij is the Kronecker delta (0 if i π j, 1 if i = j). Thus the value of dxi

on an arbitrary tangent vector vp is the ith coordinate vi of its vector part—anddoes not depend on the point of application p.

(2) The 1-form y = f1dx1 + f2dx2 + f3dx3. Since dxi is a 1-form, our def-initions show that y is also a 1-form for any functions f1, f2, f3. The valueof y on an arbitrary tangent vector vp is

The first of these examples shows that the 1-forms dx1, dx2, dx3 are the ana-logues for tangent vectors of the natural coordinate functions x1, x2, x3 forpoints. Alternatively, we can view dx1, dx2, dx3 as the “duals” of the naturalunit vector fields U1, U2, U3. In fact, it follows immediately from (1) abovethat the function dxi (Uj ) has the constant value dij.

We now show that every 1-form can be written in the concrete mannergiven in (2) above.

5.4 Lemma If f is a 1-form on R3, then f = fidxi , where fi = f (Ui).These functions f1, f2, f3 are called the Euclidean coordinate functions of f.

Proof. By definition, a 1-form is a function on tangent vectors; thus fand fidxi are equal if and only if they have the same value on everytangent vector vp = viUi(p). In (2) of Example 5.3 we saw that

On the other hand,

since fi = f(Ui). Thus f and fidxi do have the same value on everytangent vector. �

This lemma shows that a 1-form on R3 is nothing more than an expressionf dx + g dy + h dz, and such expressions are now rigorously defined as func-tions on tangent vectors. Let us now show that the definition of differentialof a function (Definition 5.2) agrees with the informal definition given at thestart of this section.

5.5 Corollary If f is a differentiable function on R3, then

dff

xdx

ii=

∂∂Â .

Â

f f fv p p pp i i i i i ivU v U v f( ) = ( )( ) = ( )( ) = ( )Â Â Â

f dx f vi i p i iÂ Â( )( ) = ( )v p .

ÂÂ

Â

y v v p v pp i i p i i p i if dx f dx f v( ) = ( )( ) = ( ) ( ) = ( )Â Â Â ,

1.5 1-Forms 25

Proof. The value of (∂f/∂xi)dxi on an arbitrary tangent vector vp is(∂f/∂xi) (p)vi. By Lemma 3.2, df(vp) = vp[ f ] is the same. Thus the 1-forms

df and (∂f/∂xi) dxi are equal. �

Using either this result or the definition of d, it is immediate that

Finally, we determine the effect of d on products of functions and on com-positions of functions.

5.6 Lemma Let fg be the product of differentiable functions f and g onR3. Then

Proof. Using Corollary 5.5, we obtain

�

5.7 Lemma Let f: R3 Æ R and h: R Æ R be differentiable functions, sothe composite function h( f ): R3 Æ R is also differentiable. Then

Proof. (The prime here is just the ordinary derivative, so h¢( f ) is again acomposite function, from R3 to R.) The usual chain rule for a compositefunction such as h( f ) reads

Hence

�

To compute df for a given function f it is almost always simpler to use theseproperties of d rather than substitute in the formula of Corollary 5.5. Then

d h fh fx

dx h ff

xdx h f df

ii

ii( )( ) =

∂ ( )( )∂

= ¢( ) ∂∂

= ¢( )Â Â .

∂ ( )( )∂

= ¢( ) ∂∂

h fx

h ff

xi i

.

d h f h f df( )( ) = ¢( ) .

d fgfgx

dxf

xg f

gx

dx

gf

xdx f

gx

dx gdf fdg

ii

i ii

ii

ii

( ) =∂( )∂

=∂∂

+∂∂

ÊËÁ

ˆ¯

=∂∂

ÊËÁ

ˆ¯

+∂∂

ÊËÁ

ˆ¯

= +

Â Â

Â Â .

d fg gdf fdg( ) = + .

d f g df dg+( ) = + .

ÂÂ

Â


from df we immediately get the partial derivatives of f, and, in fact, all itsdirectional derivatives. For example, suppose

Then by Lemmas 5.6 and 5.7,

Now use the rules above to evaluate this expression on a tangent vector vp.The result is

Exercises

1. Let v = (1, 2, -3) and p = (0, -2, 1). Evaluate the following 1-formson the tangent vector vp.

(a) y2 dx. (b) z dy - y dz.(c) (z2 - 1)dx - dy + x2 dz.

2. If f = fidxi and V = viUi, show that the 1-form f evaluated on thevector field V is the function f(V) = fivi.

3. Evaluate the 1-form f = x2 dx - y2 dz on the vector fieldsV = xU1 + yU2 + zU3,W = xy (U1 - U3) + yz (U1 - U2), and (1/x)V + (1/y)W.

4. Express the following differentials in terms of df:(a) d( f 5). (b) , where f > 0.(c) d(log(1 + f 2)).

5. Express the differentials of the following functions in the standard formfi dxi.(a) (x2 + y2 + z2)1/2. (b) tan-1(y/x).

6. In each case compute the differential of f and find the directional deriv-ative vp[ f ], for vp as in Exercise 1.

(a) f = xy2 - yz2. (b) f = xeyz.(c) f = sin(xy) cos(xz).

Â

d f( )

ÂÂÂ

v vp pf df p p v p p p v p v[ ] = ( ) = + + -( ) + +( )2 2 1 11 2 1 12

2 3 2 22

3.

df x dx y x dy y dy z y dz

xy dx x yz dy y dzf x f y f z

= ( ) + -( ) + ( ) + +( )= + + -( ) + +( )

∂ ∂ ∂ ∂ ∂ ∂

2 1 2 2

2 2 1 2

2 2

2 2

{ 1 244 344 124 34.

f x y y z= -( ) + +( )2 21 2 .

1.5 1-Forms 27

7. Which of the following are 1-forms? In each case f is the function ontangent vectors such that the value of f on (v1, v2, v3)p is

(a) v1 - v3. (b) p1 - p3.(c) v1p3 + v2p1. (d) vp[x2 + y2].(e) 0. (f) (p1)2.

In case f is a 1-form, express it as fi dxi.

8. Prove Lemma 5.6 directly from the definition of d.

9. A 1-form f is zero at a point p provided f(vp) = 0 for all tangent vectorsat p. A point at which its differential df is zero is called a critical point of thefunction f. Prove that p is a critical point of f if and only if

Find all critical points of f = (1 - x2)y + (1 - y2)z.

(Hint: Find the partial derivatives of f by computing df.)

10. (Continuation.) Prove that the local maxima and local minima of f arecritical points of f. (f has a local maximum at p if f(q) � f(p) for all q nearp.)

11. It is sometimes asserted that df is the linear approximation of Df.(a) Explain the sense in which (df )(vp) is a linear approximation off (p + v) - f(p).(b) Compute exact and approximate values of f(0.9, 1.6, 1.2) - f(1, 1.5, 1),where f = x2y/z.

1.6 Differential Forms

The 1-forms on R3 are part of a larger system called the differential forms onR3. We shall not give as rigorous an account of differential forms as we didof 1-forms since our use of the full system on R3 is limited. However, theproperties established here are valid whenever differential forms are used.

Roughly speaking, a differential form on R3 is an expression obtained byadding and multiplying real-valued functions and the differentials dx1, dx2,dx3 of the natural coordinate functions of R3. These two operations obey theusual associative and distributive laws; however, the multiplication is notcommutative. Instead, it obeys the

alternation rule: dx dx dx dx i ji j j i= - £ £( )1 3, .

∂∂

( ) =∂∂

( ) =∂∂

( ) =fx

fy

fz

p p p 0.

Â


This rule appears—although rather inconspicuously—in elementary calculus(see Exercise 9).

A consequence of the alternation rule is the fact that “repeats are zero,”that is, dxi dxi = 0, since if i = j the alternation rule reads

If each summand of a differential form contains p dxi’s (p = 0, 1, 2, 3), theform is called a p-form, and is said to have degree p. Thus, shifting to dx, dy,dz, we find

A 0-form is just a differentiable function f.A 1-form is an expression f dx + g dy + h dz, just as in the preceding section.A 2-form is an expression f dx dy + g dx dz + h dy dz.A 3-form is an expression f dx dy dz.We already know how to add 1-forms: simply add corresponding coeffi-

cient functions. Thus, in index notation,

The corresponding rule holds for 2-forms or 3-forms.On three-dimensional Euclidean space, all p-forms with p > 3 are zero. This

is a consequence of the alternation rule, for a product of more than threedxi’s must contain some dxi twice, but repeats are zero, as noted above. Forexample, dx dy dx dz = -dx dx dy dz = 0, since dx dx = 0. As a reminderthat the alternation rule is to be used, we denote this multiplication of formsby a wedge Ÿ. (However, we do not bother with the wedge when only prod-ucts of dx, dy, dz are involved.)

6.1 Example Computation of wedge products.

(1) Let

Then

But dx dx = 0 and dy dx = -dx dy. Thus

In general, the product of two 1-forms is a 2-form.

f yŸ = + -yz dx dy x dx dz xy dy dz2 .

f yŸ = -( ) Ÿ +( )= + - -

x dx y dy z dx x dz

xz dx dx x dx dz yz dy dx yx dy dz2 .

f y= - = +x dx y dy z dx x dzand .

f dx g dx f g dxi i i i i i iÂ Â Â+ = +( ) .

dx dx dx dxi i i i= - .

1.6 Differential Forms 29

(2) Let f and y be the 1-forms given above and let q = z dy. Then

Since dy dx dy and dy dy dz each contain repeats, both are zero. Thus

(3) Let f be as above, and let h be the 2-form y dx dz + x dy dz. Omittingforms containing repeats, we find

It should be clear from these examples that the wedge product of a p-formand a q-form is a ( p + q)-form. Thus such a product is automatically zerowhenever p + q > 3.

6.2 Lemma If f and y are 1-forms, then

Proof. Write

Then by the alternation rule,

�

In the language of differential forms, the operator d of Definition 5.2 con-verts a 0-form f into a 1-form df. It is easy to generalize to an operator (alsodenoted by d ) that converts a p-form h into a ( p + 1)-form dh: One simplyapplies d (of Definition 5.2) to the coefficient functions of h. For example,here is the case p = 1.

6.3 Definition If f = fi dxi is a 1-form on R3, the exterior derivativeof f is the 2-form df = dfi Ÿ dxi.

If we expand the preceding definition using Corollary 5.5, we obtain thefollowing interesting formula for the exterior derivative of

f

f

= + +

=∂∂

-∂∂

ÊËÁ

ˆ¯

+∂∂

-∂∂

ÊËÁ

ˆ¯

+∂∂

-∂∂

ÊËÁ

ˆ¯

f dx f dx f dx

dfx

fx

dx dxfx

fx

dx dxfx

fx

dx dx

1 1 2 2 3 3

2

1

1

21 2

3

1

1

31 3

3

2

2

32 3

:

.

ÂÂ

f y y fŸ = - = - ŸÂ Âf g dx dx g f dx dxi j i j j i j i= .

f y= =Â Âf dx g dxi i i i, .

f y y fŸ = - Ÿ .

f hŸ = - = +( )x dx dy dz y dy dx dz x y dx dy dz2 2 2 2 .

q f yŸ Ÿ = -x z dx dy dz2 .

q f yŸ Ÿ = + -yz dy dx dy x z dy dx dz xyz dy dy dz2 2 .


There is no need to memorize this formula; it is more reliable simply to applythe definition in each case. For example, suppose

Then

It is easy to check that the general exterior derivative enjoys the same lin-earity property as the particular case in Definition 5.2; that is,

where f and y are arbitrary forms and a and b are numbers.The exterior derivative and the wedge product work together nicely:

6.4 Theorem Let f and g be functions, f and y 1-forms. Then

(1) d( fg) = df g + f dg.(2) d( ff) = df Ÿ f + f df.(3) d(f Ÿ y) = df Ÿ y - f Ÿ dy.†

Proof. The first formula is just Lemma 5.6. We include it to show thefamily resemblance of all three formulas. The proof of (2) is a simplerversion of that of (3), so we outline a proof of the latter—watching to seewhere the minus sign comes from.

It suffices to prove the formula when f = f du, y = g dv, where u and vare any of the coordinate functions x1, x2, x3. In fact, every 1-form is a sumof such terms, so the general case will follow by the linearity of d and thealgebra of wedge products.

For example, let us try the typical case f = f dx, y = g dy. Since repeatskill, there is no use writing down terms that are bound to be eliminated.Hence

(*)d d fg dx dyfgz

dz dx dy fgz

gfz

dx dy dzf yŸ( ) = ( ) =∂( )

∂=

∂∂

+∂∂

ÊËÁ

ˆ¯

.

d a b a d b df y f y+( ) = + ,

d d xy dx d x dz

y dx x dy dx x dx dz

x dx dy x dx dz

f = ( ) Ÿ + ( ) Ÿ

= +( ) Ÿ + ( ) Ÿ

= - +

2

2

2 .

f = +xy dx x dz2 .


† As usual, multiplication takes precedence over addition or subtraction, so this expression shouldbe read as (df Ÿ y) - (f Ÿ dy).

Now,

But

since dx dz dy = -dx dy dz. Thus we must subtract this last equation fromits predecessor to get (*). �

One way to remember the minus sign in equation (3) of the theorem is totreat d as if it were a 1-form. To reach y, d must change places with f, hencethe minus sign is consistent with the alternation rule in Lemma 6.2.

Differential forms, and the associated notions of wedge product and exte-rior derivative, provide the means of expressing quite complicated relationsamong the partial derivatives in a highly efficient way. The wedge productsaves much useless labor by discarding, right at the start, terms that will even-tually disappear. But the exterior derivative d is the key. Exercise 8 shows, forexample, how it replaces all three of the differentiation operations of classi-cal vector analysis.

Exercises

1. Let f = yz dx + dz, y = sin z dx + cos z dy, x = dy + z dz. Find thestandard expressions (in terms of dxdy, . . .) for

(a) f Ÿ y, y Ÿ x, x Ÿ f. (b) df, dy, dx.

2. Let f = dx/y and y = z dy. Check the Leibnizian formula (3) ofTheorem 6.4 in this case by computing each term separately.

3. For any function f show that d(df ) = 0. Deduce that d( f dg) = df Ÿ dg.

4. Simplify the following forms:(a) d( f dg + g df ). (b) d(( f - g) (df + dg)).(c) d( f dg Ÿ g df ). (d) d(gf df) + d( f dg).

5. For any three 1-forms fi = j fijdxj (1 � i � 3), prove

f f f1 2 3

11 12 13

21 22 23

31 32 33

1 2 3Ÿ Ÿ =f f f

f f f

f f f

dx dx dx .

Â

f yŸ = Ÿ ( ) = Ÿ∂∂

Ÿ = -∂∂

d f dx d g dy f dxgz

dz dy fgz

dx dy dz,

d d f dx g dyfz

dz dx g dy gfz

dx dy dzf yŸ = ( ) Ÿ =∂∂

Ÿ Ÿ =∂∂

.


6. If r, J, z are the cylindrical coordinate functions on R3, then x = rcosJ,y = r sin J, z = z. Compute the volume element dx dy dz of R3 in cylindricalcoordinates. (That is, express dx dy dz in terms of the functions r, J, z, andtheir differentials.)

7. For a 2-form

the exterior derivative dh is defined to be the 3-form obtained by replacing f, g, and h by their differentials. Prove that for any 1-form f,d(df) = 0.

Exercises 3 and 7 show that d 2 = 0, that is, for any form x, d(dx) = 0. (Ifx is a 2-form, then d(dx) = 0, since its degree exceeds 3.)

8. Classical vector analysis avoids the use of differential forms on R3 by con-verting 1-forms and 2-forms into vector fields by means of the following one-to-one correspondences:

Vector analysis uses three basic operations based on partial differentiation:Gradient of a function f:

Curl of a vector field V = fiUi:

Divergence of a vector field V = fiUi:

Prove that all three operations may be expressed by exterior derivatives asfollows:

(a)

(b)

(c) If , then divh h´ = ( )( )1

V d V dx dy dz.

If , then curlf f´ ´( ) ( )1 2

V d V.

df f´( )1

grad .

div Vfx

i

i

=∂∂Â .

Â

curl Vfx

fx

Ufx

fx

Ufx

fx

U=∂∂

-∂∂

ÊËÁ

ˆ¯

+∂∂

-∂∂

ÊËÁ

ˆ¯

+∂∂

-∂∂

ÊËÁ

ˆ¯

3

2

2

31

1

3

3

12

2

1

1

23.

Â

grad ff

xU

ii=

∂∂Â .

f dx fU f dx dx f dx dx f dx dxi i i iÂ Â´ ´ - +( ) ( )1 2

3 1 2 2 1 3 1 2 3.

h = + +f dx dy g dx dz h dy dz,


9. Let f and g be real-valued functions on R2. Prove that

This formula appears in elementary calculus; show that it implies the alter-nation rule.

1.7 Mappings

In this section we discuss functions from Rn to Rm. If n = 3 and m = 1, thensuch a function is just a real-valued function on R3. If n = 1 and m = 3, itis a curve in R3. Although our results will necessarily be stated for arbitrarym and n, we are primarily interested in only three other cases:

The fundamental observation about a function F: Rn Æ Rm is that it canbe completely described by m real-valued functions on Rn. (We saw thisalready in Section 4 for n = 1, m = 3.)

7.1 Definition Given a function F: Rn Æ Rm, let f1, f2, . . . , fm, denotethe real-valued functions on Rn such that

for all points p in Rn. These functions are called the Euclidean coordinate func-tions of F, and we write F = (f1, f2, . . . , fm).

The function F is differentiable provided its coordinate functions are dif-ferentiable in the usual sense. A differentiable function F: Rn Æ Rm is calleda mapping from Rn to Rm.

Note that the coordinate functions of F are the composite functions fi = xi(F ), where x1, . . . , xm are the coordinate functions of Rm.

Mappings may be described in many different ways. For example, supposeF: R3 Æ R3 is the mapping F = (x2, yz, xy). Thus

Now, p = (p1, p2, p3), and by definition of the coordinate functions,

F x y z x yp p p p p p p( ) = ( ) ( ) ( ) ( ) ( )( )2, , for all .

F f f fmp p p p( ) = ( ) ( ) ( )( )1 2, , . . . ,

R R R R R R2 2 2 3 3 3Æ Æ Æ, , .

df dg

fx

fy

gx

gy

dx dyŸ =

∂∂

∂∂

∂∂

∂∂

.


Hence we obtain the following pointwise formula for F:

Thus, for example,

In principle, one could deduce the theory of curves from the general theoryof mappings. But curves are reasonably simple, while a mapping, even in thecase R2 Æ R2, can be quite complicated. Hence we reverse this process anduse curves, at every stage, to gain an understanding of mappings.

7.2 Definition If a: I Æ Rn is a curve in Rn and F: Rn Æ Rm is a mapping,then the composite function b = F(a): I Æ Rm is a curve in Rm called theimage of a under F (Fig. 1.13).

7.3 Example Mappings. (1) Consider the mapping F: R3 Æ R3 such that

In pointwise terms then,

Only when a mapping is quite simple can one hope to get a good idea of itsbehavior by merely computing its values at some finite number of points. Butthis function is quite simple —it is a linear transformation from R3 to R3.

F p p p p p p p p p p p1 2 3 1 2 1 2 3 1 2 32, , , , for all , ,( ) = - +( ) .

F x y x y z= - +( ), , 2 .

F F1 2 0 1 0 2 3 1 3 9 3 3, , , , and , ,-( ) = -( ) -( ) = -( ), , .

F p p p p p p p p p p p1 2 3 12

2 3 1 2 1 2 3, , , , for all , ,( ) = ( ) .

x p y p z pp p p( ) = ( ) = ( ) =1 2 3, , .

1.7 Mappings 35

FIG. 1.13

Thus by a well-known theorem of linear algebra, F is completely determinedby its values at three (linearly independent) points, say the unit points

(2) The mapping F: R2 Æ R2 such that F(u, v) = (u2 - v2, 2uv). (Here u andv are the coordinate functions of R2.) To analyze this mapping, we examineits effect on the curve a(t) = (rcos t, r sin t), where 0 � t � 2p. This curvetakes one counterclockwise trip around the circle of radius r with center atthe origin. The image curve is

with 0 � t � 2p. Using the trigonometric identities

we find for b = F(a) the formula

with 0 � t � 2p. This curve takes two counterclockwise trips around the circleof radius r2 centered at the origin (Fig. 1.14).

Thus the effect of F is to wrap the plane R2 smoothly around itself twice—leaving the origin fixed, since F(0, 0) = (0, 0). In this process, each circle ofradius r is wrapped twice around the circle of radius r2.

Generally speaking, differential calculus deals with approximation ofsmooth objects by linear objects. The best-known case is the approximationof a differentiable real-valued function f near x by the linear function Dx Æf ¢(x) Dx, which gives the tangent line at x to the graph of f. Our goal now isto define an analogous linear approximation for a mapping F: Rn Æ Rm neara point p of Rn.

b t r t r t( ) = ( )2 22 2cos sin, ,

cos cos sin sin sin cos2 2 22 2t t t t t t= - =, ,

b at F t

F r t r t

r t r t r t t

( ) = ( )( )= ( )= -( )

cos sin

cos sin cos sin

,

,2 2 2 2 22

u u u1 2 31 0 0 0 1 0 0 0 1= ( ) = ( ) = ( ), , , , , , ,, .


R2 R2

r2

vF

u ur

a b

v

FIG. 1.14

Since Rn is filled by the radial lines a(t) = p + tv starting at p, Rm is filledby their image curves b(t) = F(p + tv) starting at F(p) (Fig. 1.15). So we approximate F near p by the map F* that sends each initial velocity a¢(0) = vp to the initial velocity b¢(0).

7.4 Definition Let F: Rn Æ Rm be a mapping. If v is a tangent vector toRn at p, let F*(v) be the initial velocity of the curve t Æ F(p + tv). The result-ing function F* sends tangent vectors to Rn to tangent vectors to Rm, and iscalled the tangent map of F.

The tangent map can be described explicitly as follows.

7.5 Proposition Let F = (f1, f2, . . . , fm) be a mapping from Rn to Rm.If v is a tangent vector to Rn at p, then

Proof. For definiteness, take m = 3. Then

By definition, F*(v) = b¢(0). To get b¢(0), we take the derivatives, at t = 0,of the coordinate functions of b (Definition 4.3). But

Thus

F f f f*( ) = [ ] [ ] [ ]( ) ( )v v v v1 2 3 0, , ,b

ddt

f t fi t ip v v+( )( ) = [ ]=0 .

b t F t f t f t f t( ) = +( ) = +( ) +( ) +( )( )p v p v p v p v1 2 3, , .

F f f Fm*( ) = [ ] [ ]( ) ( )v v v p1 , . . . , .at

1.7 Mappings 37

FIG. 1.15

and b(0) = F(p). �

Fix a point p of Rn. The definition of tangent map shows that F* sendstangent vectors at p to tangent vectors at F(p). Thus for each p in Rn, thefunction F* gives rise to a function

called the tangent map of F at p. (Compare the analogous situation in elementary calculus where a function f: R Æ R has a derivative function f ¢: R Æ R that at each point t of R gives the derivative of f at t.)

7.6 Corollary If F: Rn Æ Rm is a mapping, then at each point p of Rn thetangent map F*p: Tp(Rn) Æ TF(p)(Rm) is a linear transformation.

Proof. We must show that for tangent vectors v and w at p and numbersa, b,

This follows immediately from the preceding proposition by using the lin-earity in assertion (1) of Theorem 3.3. �

In fact, the tangent map F*p at p is the linear transformation that bestapproximates F near p. This idea is fully developed in advanced calculus,where it is used to prove Theorem 7.10.

Another consequence of the proposition is that mappings preserve veloci-ties of curves. Explicitly:

7.7 Corollary Let F: Rn Æ Rm be a mapping. If b = F(a) is the image ofa curve a in Rn, then b¢ = F*(a ¢).

Proof. Again, set m = 3. If F = (f1, f2, f3), then

Hence Theorem 7.5 gives

But by Lemma 4.6,

F f f f* ¢( ) = ¢[ ] ¢[ ] ¢[ ]( )a a a a1 2 3, .,

b a a a a= ( ) = ( ) ( ) ( )( )F f f f1 2 3, , .

F a b aF bF* +( ) = *( ) + *( )v w v w .

F T Tp pn

F pm

* ( ) Æ ( )( ): R R


Hence

�

Let {Uj} (1 £ j £ n) and {Ui} (1 £ i £ m) be the natural frame fields of Rn

and Rm, respectively (Def. 2.4). Then:

7.8 Corollary If F = (f1, . . . , fm) is a mapping from Rn to Rm, then

Proof. This follows directly from Proposition 7.5, since .

�

The matrix appearing in the preceding formula,

is called the Jacobian matrix of F at p. (When m = n = 1; it reduces to asingle number: the derivative of F at p.)

Just as the derivative of a function is used to gain information about thefunction, the tangent map F* can be used in the study of a mapping F.

7.9 Definition A mapping F: Rn Æ Rm is regular provided that at everypoint p of Rn the tangent map F*p is one-to-one.

Since tangent maps are linear transformations, standard results of linearalgebra show that the following conditions are equivalent:

(1) F*p is one-to-one.(2) F*(vp) = 0 implies vp = 0.(3) The Jacobian matrix of F at p has rank n, the dimension of the domain

Rn of F.

∂∂

( )ÊËÁ

ˆ¯

£ £ £ £

fx

i

j i m j n

p1 1,

U ff

xj ii

j

[ ] =∂∂

F Uf

xU F j nj

i

ji

i

m

* ( )( ) =∂∂

( ) ( )( ) £ £( )=

Âp p p1

1 .

F tdf

dtt

dfdt

tdf

dtt t

t* ¢( )( ) =

( ) ( ) ( ) ( ) ( ) ( )ÊË

ˆ¯ = ¢( )

( )a

a a ab

b

1 2 3, , .

¢[ ] =( )

aa

fdf

dtii .

1.7 Mappings 39

The following noteworthy property of linear transformations T: V Æ Wwill be useful in dealing with tangent maps. If the vector spaces V and Whave the same dimension, then T is one-to-one if and only if it is onto, soeither property is equivalent to T being a linear isomorphism.

A mapping that has a (differentiable) inverse mapping is called a diffeo-morphism. The results of this section all remain valid when Euclidean spacesRn are replaced by open sets of Euclidean spaces, so we can speak of a dif-feomorphism from one open set to another.

We state without proof one of the fundamental results of advanced cal-culus.

7.10 Theorem Let F: Rn Æ Rn be a mapping between Euclidean spacesof the same dimension. If F*p is one-to-one at a point p, there is an open setU containing p such that F restricted to U is a diffeomorphism of U onto anopen set V.

This is called the inverse function theorem since it asserts that the restrictedmapping U ÆV has a differentiable inverse mapping V Æ U. Exercise 6 givesa suggestion of its importance.

Exercises

In the first four exercises F denotes the mapping F(u, v) = (u2 - v2, 2uv) inExample 7.3.

1. Find all points p such that(a) F(p) = (0, 0). (b) F(p) = (8, 6).(c) F(p) = p.

2. (a) Sketch the horizontal line v = 1 and its image under F (a parabola).(b) Do the same for the vertical u = 1.(c) Describe the image of the unit square 0 � u, v � 1 under F.

3. Let v = (v1, v2) be a tangent vector to R2 at p = (p1, p2). Apply Defini-tion 7.4 directly to express F*(v) in terms of the coordinates of v and p.

4. Find a formula for the Jacobian matrix of F at all points, and deducethat F*p is a linear isomorphism at every point of R2 except the origin.

5. If F: Rn Æ Rm is a linear transformation, prove that F*(vp) = F(v)F(p).

6. (a) Give an example to demonstrate that a one-to-one and ontomapping need not be a diffeomorphism. (Hint: Take m = n = 1.)


(b) Prove that if a one-to-one and onto mapping F: Rn Æ Rn is regular,then it is a diffeomorphism.

7. Prove that a mapping F: Rn Æ Rm preserves directional derivatives in thissense: If vp is a tangent vector to Rn and g is a differentiable function on Rm,then F*(vp)[g] = vp[g(F )].

8. In the definition of tangent map (Def. 7.4), the straight line t Æ p + tvcan be replaced by any curve a with initial velocity vp.

9. Let F: Rn Æ Rm and G: Rm Æ Rp be mappings. Prove:(a) Their composition GF: Rn Æ Rp is a (differentiable) mapping. (Take m = p = 2 for simplicity.)(b) (GF )* = G*F*. (Hint: Use the preceding exercise.)This concise formula is the general chain rule. Unless dimensions are small,it becomes formidable when expressed in terms of Jacobian matrices.(c) If F is a diffeomorphism, then so is its inverse mapping F -1.

10. Show (in two ways) that the map F: R2 Æ R2 such that F(u, v) =(veu, 2u) is a diffeomorphism:

(a) Prove that it is one-to-one, onto, and regular;(b) Find a formula for its inverse F -1: R2 Æ R2 and observe that F -1 is dif-ferentiable. Verify the formula by checking that both F F -1 and F -1 F areidentity maps.

1.8 Summary

Starting from the familiar notion of real-valued functions and using linearalgebra at every stage, we have constructed a variety of mathematical objects.The basic notion of tangent vector led to vector fields, which dualized to 1-forms—which in turn led to arbitrary differential forms. The notions of curve and differentiable function were generalized to that of a mapping F: Rn Æ Rm.

Then, starting from the usual notion of the derivative of a real-valued func-tion, we proceeded to construct appropriate differentiation operations forthese objects: the directional derivative of a function, the exterior derivativeof a form, the velocity of a curve, the tangent map of a mapping. These oper-ations all reduced to (ordinary or partial) derivatives of real-valued coordi-nate functions, but it is noteworthy that in most cases the definitions of theseoperations did not involve coordinates. (This could be achieved in all cases.)Generally speaking, these differentiation operations all exhibited in one form

1.8 Summary 41

or another the characteristic linear and Leibnizian properties of ordinary differentiation.

Most of these concepts are probably already familiar to the reader, at least in special cases. But we now have careful definitions and a catalogue ofbasic properties that will enable us to begin the exploration of differentialgeometry.


▼

▲Chapter 2

Frame Fields

43

Roughly speaking, geometry begins with the measurement of distances andangles. We shall see that the geometry of Euclidean space can be derived fromthe dot product, the natural inner product on Euclidean space.

Much of this chapter is devoted to the geometry of curves in R3. Weemphasize this topic not only because of its intrinsic importance, but alsobecause the basic method used to investigate curves has proved effectivethroughout differential geometry. A curve in R3 is studied by assigning at eachpoint a certain frame—that is, set of three orthogonal unit vectors. The rateof change of these vectors along the curve is then expressed in terms of thevectors themselves by the celebrated Frenet formulas (Theorem 3.2). In a realsense, the theory of curves in R3 is merely a corollary of these fundamentalformulas.

Later on we shall use this “method of moving frames” to study a surfacein R3. The general idea is to think of a surface as a kind of two-dimensionalcurve and follow the Frenet approach as closely as possible. To carry out thisscheme we shall need the generalization (Theorem 7.2) of the Frenet formu-las devised by E. Cartan. It was Cartan who, in the early 1900s, first realizedthe full power of this method not only in differential geometry but also in avariety of related fields.

2.1 Dot Product

We begin by reviewing some basic facts about the natural inner product onthe vector space R3.

1.1 Definition The dot product of points p = ( p1, p2, p3) and q = (q1, q2,q3) in R3 is the number

The dot product is an inner product since it has the following three properties:

(1) Bilinearity:

(2) Symmetry: p • q = q • p.(3) Positive definiteness: p • p � 0, and p • p = 0 if and only if p = 0.(Here p, q, and r are arbitrary points of R3, and a and b are numbers.)

The norm of a point p = ( p1, p2, p3) is the number

The norm is thus a real-valued function on R3; it has the fundamental properties � p + q � � � p � + � q � and � ap � = | a | � p �, where | a | is theabsolute value of the number a.

In terms of the norm we get a compact version of the usual distanceformula in R3.

1.2 Definition If p and q are points of R3, the Euclidean distance from p to q is the number

In fact, since

expansion of the norm gives the well-known formula (Fig. 2.1)

Euclidean distance may be used to give a more precise definition of opensets (Chapter 1, Section 1). First, if p is a point of R3 and e > 0 is a number,the e neighborhood Ne of p in R3 is the set of all points q of R3 such that d(p,q) < e. Then a subset O of R3 is open provided that each point of O has an eneighborhood that is entirely contained in O. In short, all points near enoughto a point of an open set are also in the set. This definition is valid with R3

replaced by Rn—or indeed any set furnished with a reasonable distance function.

d p q p q p qp q,( ) = -( ) + -( ) + -( )( )1 12

2 22

3 32 1 2

.

p q- = - - -( )p q p q p q1 1 2 2 3 3, , ,

d p q p q,( ) = - .

p p p= ( ) = + +( )• .1 2

12

22

32 1 2

p p p

a b a b

a b a b

p q r p r q r

r p q r p r q

+( ) = +

+( ) = +

• • •

• • • .

,

p q• .= + +p q p q p q1 1 2 2 3 3

44 2. Frame Fields

We saw in Chapter 1 that for each point p of R3 there is a canonical iso-morphism v Æ vp from R3 onto the tangent space Tp(R3) at p. These isomor-phisms lie at the heart of Euclidean geometry—using them, the dot producton R3 itself may be transferred to each of its tangent spaces.

1.3 Definition The dot product of tangent vectors vp and wp at the samepoint of R3 is the number vp • wp = v • w.

For example, (1, 0, -1)p • (3, -3, 7)p = 1(3) + 0(-3) + (-1)7 = -4. Evidentlythis definition provides a dot product on each tangent space Tp(R3) with thesame properties as the original dot product on R3. In particular, each tangentvector vp to R3 has norm (or length) � vp � = � v �.

A fundamental result of linear algebra is the Schwarz inequality | v • w |� � v � � w �. This permits us to define the cosine of the angle J between vand w by the equation (Fig. 2.2).

Thus the dot product of two vectors is the product of their lengths times thecosine of the angle between them. (The angle J is not uniquely determinedunless further restrictions are imposed, say 0 � J � p.)

In particular, if J = p/2, then v • w = 0. Thus we shall define two vectorsto be orthogonal provided their dot product is zero. A vector of length 1 iscalled a unit vector.

1.4 Definition A set e1, e2, e3 of three mutually orthogonal unit vectorstangent to R3 at p is called a frame at the point p.

Thus e1, e2, e3 is a frame if and only if

e e e e e e

e e e e e e

1 1 2 2 3 3

1 2 1 3 2 3

1

0

• • •

• • • .

= = =

= = =

,

v w v w• cos .= J

2.1 Dot Product 45

FIG. 2.1 FIG. 2.2

By the symmetry of the dot product, the second row of equations is, ofcourse, the same as

Using index notation, all nine equations may be concisely expressed as ei • ej = dij for 1 � i j � 3, where dij is the Kronecker delta (0 if i π j, 1 ifi = j). For example, at each point p of R3, the vectors U1(p), U2(p), U3(p) ofDefinition 2.4 in Chapter 1 constitute a frame at p.

1.5 Theorem Let e1, e2, e3 be a frame at a point p of R3. If v is any tangentvector to R3 at p, then (Fig. 2.3)

Proof. First we show that the vectors e1, e2, e3 are linearly independent.Suppose aiei = 0. Then

where all sums are over i = 1, 2, 3. Thus

as required. Now, the tangent space Tp(R3) has dimension 3, since it is lin-early isomorphic to R3. Thus by a well-known theorem of linear algebra,the three independent vectors e1, e2, e3 form a basis for Tp(R3). Hence foreach vector v there are three (unique) numbers c1, c2, c3 such that

But

v e e e• •j i i j i ij jc c c= ( ) = =Â Â d ,

v e= Â ci i .

a a a1 2 3 0= = = ,

0 = ( ) = = =Â Â Âa a a ai i j i i j i ij je e e e• • d ,

Â

v v e e v e e v e e= ( ) + ( ) + ( )• • • .1 1 2 2 3 3

e e e e e e2 1 3 1 3 2 0• • • .= = =

46 2. Frame Fields

FIG. 2.3

and thus

◆

This result (valid in any inner-product space) is one of the great labor-saving devices in mathematics. For to find the coordinates of a vector v withrespect to an arbitrary basis, one must in general solve a set of nonhomoge-neous linear equations, a task that even in dimension 3 is not always entirelytrivial. But the theorem shows that to find the coordinates of v with respectto a frame (that is, an orthonormal basis) it suffices merely to compute thethree dot products v • e1, v • e2, v • e3. We call this process orthonormal expan-sion of v in terms of the frame e1, e2, e3. In the special case of the naturalframe U1(p), U2(p), U3(p), the identity

is an orthonormal expansion, and the dot product is defined in terms of theseEuclidean coordinates by If we use instead an arbitrary framee1, e2, e3, then each vector v has new coordinates ai = v • ei relative to thisframe, but the dot product is still given by the same simple formula

since

When applied to more complicated geometric situations, the advantage ofusing frames becomes enormous, and this is why they appear so frequentlythroughout this book.

The notion of frame is very close to that of orthogonal matrix.

1.6 Definition Let e1, e2, e3 be a frame at a point p of R3. The 3 ¥ 3 matrixA whose rows are the Euclidean coordinates of these three vectors is calledthe attitude matrix of the frame.

Explicitly, if

e

e

e

1 11 12 13

2 21 22 23

3 31 32 33

= ( )= ( )= ( )

a a a

a a a

a a a

p

p

p

, , ,

, , ,

, , ,

v w e e e e• • •

.

,

,

= ( ) ( ) =

= =

Â Â Â

Â Â

a b a b

a b a b

i i i i i j i ji j

i j iji j

i id

v w• = Âa bi i

v w• .= Â v wi i

v p= ( ) = ( )Âv v v vUi i1 2 3, ,

v v e e= ( )Â • .j j

2.1 Dot Product 47

then

Thus A does describe the “attitude” of the frame in R3, although not its pointof application.

Evidently the rows of A are orthonormal, since

By definition, this means that A is an orthogonal matrix.In terms of matrix multiplication, these equations may be written

A tA = I, where I is the 3 ¥ 3 identity matrix and tA is the transpose of A:

It follows by a standard theorem of linear algebra that tAA = I, so that tA = A-1, the inverse of A.

There is another product on R3, closely related to the wedge product of 1-forms and second in importance only to the dot product. We shall transfer itimmediately to each tangent space of R3.

1.7 Definition If v and w are tangent vectors to R3 at the same point p,then the cross product of v and w is the tangent vector

This formal determinant is to be expanded along its first row. For example,if v = (1, 0, -1)p and w = (2, 2, -7)p, then

Familiar properties of determinants show that the cross product v ¥ w islinear in v and in w, and satisfies the alternation rule

v w w v¥ = - ¥ .

v w

p p p

p p p

¥ =( ) ( ) ( )

--

= ( ) + ( ) + ( ) = ( )

U U U

U U Up

1 2 3

1 2 3

1 0 1

2 2 7

2 5 2 2 5 2, , .

v w

p p p

¥ =( ) ( ) ( )U U U

v v v

w w w

1 2 3

1 2 3

1 2 3

.

tA

a a a

a a a

a a a

=Ê

Ë

ÁÁ

ˆ

¯

˜˜

11 21 31

12 22 32

13 23 33

.

a a i jik jkk i j ijÂ = = £ £e e• .d for ,1 3

A

a a a

a a a

a a a

=Ê

Ë

ÁÁ

ˆ

¯

˜˜

11 12 13

21 22 23

31 32 33

.

48 2. Frame Fields

Hence, in particular, v ¥ v = 0. The geometric usefulness of the cross productis based mostly on this fact:

1.8 Lemma The cross product v ¥ w is orthogonal to both v and w, andhas length such that

Proof. Let Then the dot product v • (v ¥ w) is justBut by the definition of cross product, the Euclidean coordinates

c1, c2, c3 of v ¥ w are such that

This determinant is zero, since two of its rows are the same; thus v ¥ w isorthogonal to v, and similarly, to w.

Rather than use tricks to prove the length formula, we give a brute-forcecomputation. Now,

On the other hand,

and expanding these squares gives the same result as above. ◆

A more intuitive description of the length of a cross product is

where 0 � J � p is the smaller of the two angles from v to w. The directionof v ¥ w on the line orthogonal to v and w is given, for practical purposes,by this “right-hand rule”: If the fingers of the right hand point in the

v w v w¥ = sin J,

v w v w v w¥ = ¥( ) ¥( ) =

= -( ) + -( ) + -( )Â2 2

2 3 3 22

3 1 1 32

1 2 2 12

• c

v w v w v w v w v w v w

i

,

v v w w v w• • •

.

( ) ( ) - ( ) = ( )( ) - ( )

= - +ÏÌÓ

¸˝˛

= -

Â Â Â

Â Â Â

Â Â<

π <

2 2 2 2

2 2 2 2

2 2

2

2

v w v w

v w v w v w v w

v w v w v w

i j i i

i ji j

i i i i j ji j

i ji j

i i j ji j

,

v v w• .¥( ) =v v v

v v v

w w w

1 2 3

1 2 3

1 2 3

v ci i .Âv w p¥ = ( )Â cUi i .

v w v v w w v w¥ = ( ) ( ) - ( )2 2• • • .

2.1 Dot Product 49

direction of the shortest rotation of v to w, then the thumb points in thedirection of v ¥ w (Fig. 2.4).

Combining the dot and cross product, we get the triple scalar product,which assigns to any three vectors u, v, w the number u • v ¥ w (Exercise 4).Parentheses are unnecessary: u • (v ¥ w) is the only possible meaning.

Exercises

1. Let v = (1, 2, -1) and w = (-1, 0, 3) be tangent vectors at a point of R3.Compute:

(a) v • w. (b) v ¥ w.(c) v/� v �, w/� w �. (d) � v ¥ w �.(e) the cosine of the angle between v and w.

2. Prove that Euclidean distance has the properties(a) d(p, q) � 0; d(p, q) = 0 if and only if p = q,(b) d(p, q) = d(q, p),(c) d(p, q) + d(q, r) � d(p, r), for any points p, q, r in R3.

3. Prove that the tangent vectors

constitute a frame. Express v = (6, 1, -1) as a linear combination of thesevectors. (Check the result by direct computation.)

4. Let u = (u1, u2, u3), v = (v1, v2, v3), w = (w1, w2, w3). Prove that

(a) u v w• .¥ =u u u

v v v

w w w

1 2 3

1 2 3

1 2 3

e e e1 2 3

1 2 1

6

2 0 2

8

1 1 1

3=

( )=

-( )=

-( ), ,,

, ,,

, ,

50 2. Frame Fields

FIG. 2.4

(b) u • v ¥ w π 0 if and only if u, v, and w are linearly independent.(c) If any two vectors in u • v ¥ w are reversed, the product changes sign.(d) u • v ¥ w = u ¥ v • w.

5. Prove that v ¥ w π 0 if and only if v and w are linearly independent, andshow that � v ¥ w � is the area of the parallelogram with sides v and w.

6. If e1, e2, e3 is a frame, show that

Deduce that any 3¥3 orthogonal matrix has determinant ±1.

7. If u is a unit vector, then the component of v in the u direction is

Show that v has a unique expression v = v1 + v2, where v1 • v2 = 0 and v1 isthe component of v in the u direction.

8. Prove: The volume of the parallelepiped with sides u, v, w is ±u • v ¥ w(Fig. 2.5). (Hint: Use the indicated unit vector e = v ¥ w/� v ¥ w �.)

9. Prove, using e-neighborhoods, that each of the following subsets of R3

is open:(a) All points p such that � p � < 1.(b) All p such that p3 > 0. (Hint: | pi - qi | � d(p, q).)

10. In each case, let S be the set of all points p that satisfy the given con-dition. Describe S, and decide whether it is open.

(a) p12 + p2

2 + p32 = 1. (b) p3 π 0.

(c) p1 = p2 π p3. (d) p12 + p2

2 < 9.

11. If f is a differentiable function on R3, show that the gradient

— =∂∂Âf

fx

Ui

i

v u u v u• cos .( ) = J

e e e1 2 3 1• .¥ = ±

2.1 Dot Product 51

FIG. 2.5

(Ex. 8 of Sec. 1.6) has the following properties:(a) v [ f ] = (df )(v) = v • (—f )(p) for any tangent vector at p.

(b) The norm of (—f ) (p) is the maximumof the directional derivatives u[ f ] for all unit vectors at p. Furthermore, if(—f )(p) π 0, the unit vector for which the maximum occurs is

The notations grad f, curl V, and div V (in the exercise referred to) areoften replaced by —f, — ¥ V, and — • V, respectively.

12. Angle functions. Let f and g be differentiable real-valued functions onan interval I. Suppose that f 2 + g2 = 1 and that J0 is a number such that f(0) = cosJ0, g(0) = sinJ0. If J is the function such that

prove that

Hint: We want ( f - cosJ)2 + (g - sinJ)2 = 0, so show that its derivative iszero.

The point of this exercise is that J is a differentiable function, unambigu-ously defined on the whole interval I.

2.2 Curves

We begin the geometric study of curves by reviewing some familiar defini-tions. Let a: I Æ R3 be a curve. In Chapter 1, Section 4, we defined the veloc-ity vector a ¢(t) of a at t. Now we define the speed of a at t to be the lengthv(t) = � a ¢(t) � of the velocity vector. Thus speed is a real-valued function onthe interval I. In terms of Euclidean coordinates a = (a1, a2, a3), we have

Hence the speed function v of a is given by the usual formula

In physics, the distance traveled by a moving point is determined by inte-grating its speed with respect to time. Thus we define the arc length of a fromt = a to t = b to be the number

vddt

ddt

ddt

= ¢ = ÊË

ˆ¯ + Ê

Ëˆ¯ + Ê

Ëˆ¯

ÊËÁ

ˆ¯

aa a a1

22

23

2 1 2

.

¢( ) = ( ) ( ) ( )ÊË

ˆ¯a

a a at

ddt

tddt

tddt

t1 2 3, , .

f g= =cos sin .J J,

J Jt fg gf dut

( ) = + ¢ - ¢( )Ú00

,

—( )( ) —( )( )f fp p .

—( )( ) = ∂ ∂( ) ( )[ ]Âf f xip p2 1 2

52 2. Frame Fields

Substituting the formula for � a¢ � given above, we get the usual formulafor arc length. This length involves only the restriction of a (defined on some open interval) to the closed interval [a, b]: a � t � b. Such a restrictions: [a, b] Æ R3 is called a curve segment, and its length is denoted by L(s). Note that the velocity of s is well defined at the endpoints a and b of[a, b].

Sometimes one is interested only in the route followed by a curve and not in the particular speed at which it traverses its route. One way to ignore the speed of a curve a is to reparametrize to a curve b that has unit speed � b¢ � = 1. Then b represents a “standard trip” along the route of a.

2.1 Theorem If a is a regular curve in R3, then there exists a reparame-trization b of a such that b has unit speed.

Proof. Fix a number a in the domain I of a: I Æ R3, and consider thearc length function

(The resulting reparametrization is said to be based at t = a.) Thus thederivative of the function s = s(t) is the speed function v = � a ¢ � ofa. Since a is regular, by definition a ¢ is never zero; hence > 0. By a standard theorem of calculus, the function s has an inverse function t = t(s), whose derivative at s = s(t) is the reciprocal of at t = t(s). In particular, > 0.

Now let b be the reparametrization b(s) = a(t(s)) of a. We assert thatb has unit speed. In fact, by Lemma 4.5 of Chapter 1,

Hence, by the preceding remarks, the speed of b is

◆

We shall use the notation of this proof frequently in later work. The unit-speed curve b is sometimes said to have arc-length parametrization, since thearc length of b from s = a to s = b (a < b) is just b - a.

¢( ) = ( ) ¢ ( )( ) = ( ) ( )( ) =b asdtds

s t sdtds

sdsdt

t s 1.

¢( ) = ( ) ¢ ( )( )b asdtds

s t s .

dt ds/ds dt/dt ds/

ds dt/ds dt/

s t u dua

b

( ) = ¢( )Ú a .

¢( )Ú a t dta

b

.

2.2 Curves 53

For example, consider the helix a in Example 4.2 of Chapter 1. Since a(t) = (acos t, asin t, bt), the velocity a ¢ is given by the formula

Hence

Thus a has constant speed c = � a ¢ � = (a2 + b2)1/2. If we measure arc lengthfrom t = 0, then

Hence, t(s) = s/c. Substituting in the formula for a, we get the unit-speed reparametrization

It is easy to check directly that � b¢(s) � = 1 for all s.A reparametrization a(h) of a curve a is orientation-preserving if h¢ ≥ 0 and

orientation-reversing if h¢ £ 0. In the latter case, a(h) still follows the route ofa but in the opposite direction. By definition, a unit-speed reparametrizationis always orientation-preserving since > 0 for a regular curve.

In the theory of curves we will frequently reparametrize regular curves toobtain unit speed; however, it is rarely possible to do this in practice. Theproblem is basic calculus: Even when the coordinate functions of the curveare rather simple, the speed function cannot usually be integrated explicitly—at least in terms of familiar functions.

The general notion of vector field (Definition 2.3 of Chapter 1) can beadapted to curves as follows.

2.2 Definition A vector field Y on curve a: I Æ R3 is a function that assigns to each number t in I a tangent vector Y(t) to R3 at the pointa(t).

We have already met such vector fields: For any curve a, its velocity a ¢ evi-dently satisfies this definition. Note that unlike a ¢, arbitrary vector fields ona need not be tangent to a, but may point in any direction (Fig. 2.6).

The properties of vector fields on curves are analogous to those of vectorfields on R3. For example, if Y is a vector field on a: I Æ R3, then for each tin I we can write

ds dt/

b assc

asc

asc

bsc

( ) = ÊË

ˆ¯ = Ê

Ëˆ¯cos sin ., ,

s t c du ctt

( ) = =Ú0.

¢( ) = ¢( ) ¢( ) = + + = +a a at t t a t a t b a b2 2 2 2 2 2 2 2• sin cos .

¢( ) = -( )a t a t a t bsin cos ., ,

54 2. Frame Fields

We have thus defined real-valued functions y1, y2, y3 on I, called the Euclid-ean coordinate functions of Y. These will always be assumed to be differen-tiable. Note that the composite function t Æ Ui (a(t)) is a vector field on a.Where it seems safe to do so, we shall often write merely Ui instead of Ui(a(t)).

The operations of addition, scalar multiplication, dot product, and crossproduct of vector fields (on the same curve) are all defined in the usual point-wise fashion. Thus if

and f(t) = (t + 1)/t, we obtain the vector fields

and the real-valued function

To differentiate a vector field on a one simply differentiates its Euclideancoordinate functions, thus obtaining a new vector field on a. Explicitly, if

then Thus, for Y as above, we get

¢ = - ¢¢ = ¢¢¢ =Y tU U Y U Y2 2 01 3 1, , and .

¢ = ÂYdydt

Uii .Y yUi i= Â ,

Y Z t t• .( )( ) = - 2

Y Z t t U t U

fY t t t U t U

Y Z t

U U U

t t

t t

t t U t U t t U

+( )( ) = + -( )( )( ) = +( ) - +( )

¥( )( ) = --

= -( ) - + -( )

21

22

1 3

1 2 3

2

2

21

32

2 23

1

1 1

0

0 1

1 1

,

,

Y t t U tU Z t t U tU( ) = - ( ) = -( ) +21 3

22 31, ,

Y t y t y t y t y t U tt i i( ) = ( ) ( ) ( )( ) = ( ) ( )( )( ) Â1 2 3, , a a .

2.2 Curves 55

FIG. 2.6

In particular, the derivative a≤ of the velocity a ¢ of a is called the accelera-tion of a. Thus if a = (a1, a2, a3), the acceleration a≤ is the vector field

on a. By contrast with velocity, acceleration is generally not tangent to thecurve.

As we mentioned earlier, in whatever form it appears, differentiation alwayshas suitable linearity and Leibnizian properties. In the case of vector fieldson a curve, it is easy to prove the linearity property

(a and b numbers) and the Leibnizian properties

If the function Y • Z is constant, the last formula shows that

This observation will be used frequently in later work. In particular, if Y hasconstant length � Y �, then Y and Y ¢ are orthogonal at each point, since � Y �2 = Y • Y constant implies 2Y • Y ¢ = 0.

Recall that tangent vectors are parallel if they have the same vector parts.We say that a vector field Y on a curve is parallel provided all its (tangent vector)values are parallel. In this case, if the common vector part is (c1, c2, c3), then

Thus parallelism for a vector field is equivalent to the constancy of its Euclidean coordinate functions.

Vanishing of derivatives is always important in calculus; here are threesimple cases.

2.3 Lemma (1) A curve a is constant if and only if its velocity is zero,a ¢ = 0.

(2) A nonconstant curve a is a straight line if and only if its accelerationis zero, a≤ = 0.

(3) A vector field Y on a curve is parallel if and only if its derivative iszero, Y ¢ = 0.

Proof. In each case it suffices to look at the Euclidean coordinate func-tions. For example, we shall prove (2). If a = (a1, a2, a3), then

Y t c c c cU tt i i( ) = ( ) =( ) Â1 2 3, , for alla .

¢ + ¢ =Y Z Y Z• • .0

fYdfdt

Y fY Y Z Y Z Y Z( )¢ = + ¢ ( )¢ = ¢ + ¢and • • • .

aY bZ aY bZ+( )¢ = ¢ + ¢

¢¢ = ÊË

ˆ¯a

a a a

a

ddt

ddt

ddt

21

2

22

2

23

2, ,

56 2. Frame Fields

Thus a≤ = 0 if and only if each . By elementary calculus, thisis equivalent to the existence of constants pi and qi such that

Thus a(t) = p + tq, and a is a straight line as defined in Example 4.2 ofChapter 1. (Note that nonconstancy implies q π 0.) ◆

Exercises

1. For the curve a(t) = (2t, t2, t3/3),(a) find the velocity, speed, and acceleration for arbitrary t, and at t = 1;(b) find the arc length function s = s(t) (based at t = 0), and determinethe arc length of a from t = -1 to t = +1.

2. Show that a curve has constant speed if and only if its acceleration iseverywhere orthogonal to its velocity.

3. Show that the curve a(t) = (cosh t, sinh t, t) has arc length functionsinh t, and find a unit-speed reparametrization of a.

4. Consider the curve a(t) = (2t, t2, log t) on I: t > 0. Show that this curvepasses through the points p = (2, 1, 0) and q = (4, 4, log2), and find its arclength between these points.

5. Suppose that b1 and b2 are unit-speed reparametrizations of the samecurve a. Show that there is a number s0 such that b2(s) = b1(s + s0) for all s.What is the geometric significance of s0?

6. Let Y be a vector field on the helix a(t) = (cos t, sin t, t). In each of the following cases, express Y in the form yiUi:

(a) Y(t) is the vector from a(t) to the origin of R3.(b) Y(t) = a ¢(t) - a≤(t).(c) Y(t) has unit length and is orthogonal to both a ¢(t) and a ≤(t).(d) Y(t) is the vector from a(t) to a(t + p).

7. A reparametrization a(h): [c, d ] Æ R3 of a curve segment a: [a, b] Æ R3

is monotone provided either

(i) h¢ ≥ 0, h(c) = a, h(d) = b or (ii) h¢ � 0, h(c) = b, h(d) = a.

Prove that monotone reparametrization does not change arc length.

Â

s t( ) = 2

a i i it p tq i( ) = + =, for , ,1 2 3.

d dti2 2 0a =

¢¢ = ÊË

ˆ¯a

a a addt

ddt

ddt

21

2

22

2

23

2, , .

2.2 Curves 57

8. Let Y be a vector field on a curve a. If a(h) is a reparametrization of a,show that the reparametrization Y(h) is a vector field on a(h), and prove thechain rule Y(h)¢ = h¢Y ¢(h).

9. (Numerical integration.) The curve segments

defined on 0 � t � p, run from the origin 0 to (0, p2, 0). Which is shorter?(See Integration in the Appendix.)

10. Let a, b: I Æ R3 be curves such that a ¢(t) and b¢(t) are parallel (sameEuclidean coordinates) at each t. Prove that a and b are parallel in the sensethat there is a point p in R3 such that b(t) = a(t) + p for all t.

11. Prove that a straight line is the shortest distance between two points inR3. Use the following scheme; let a: [a, b] Æ R3 be an arbitrary curve segmentfrom p = a(a) to q = a(b). Let u = (q - p)/� q - p �.

(a) If s is a straight line segment from p to q, say

show that L(s) = d(p, q).(b) From � a ¢ � � a ¢ • u, deduce L(a) � d(p, q), where L(a) is the lengthof a and d is Euclidean distance.(c) Furthermore, show that if L(a) = d(p, q), then (but for parametriza-tion) a is a straight line segment. (Hint: write a ¢ = (a ¢ • u)u + Y, where Y • u = 0.)

2.3 The Frenet Formulas

We now derive mathematical measurements of the turning and twisting of acurve in R3. Throughout this section we deal only with unit-speed curves; inthe next we extend the results to arbitrary regular curves.

Let b: I Æ R3 be a unit-speed curve, so � b ¢ (s) � = 1 for each s in I.Then T = b¢ is called the unit tangent vector field on b. Since T has constantlength 1, its derivative T ¢ = b≤ measures the way the curve is turning in R3.We call T ¢ the curvature vector field of b. Differentiation of T • T = 1 gives2T ¢ • T = 0, so T ¢ is always orthogonal to T, that is, normal to b.

The length of the curvature vector field T ¢ gives a numerical measurementof the turning of b. The real-valued function k such that k(s) = � T ¢ (s) � for

s t t t t( ) = -( ) + ( )1 0 1p q � � ,

a bt t t t t t t t t t t( ) = ( ) ( ) = +( )( )sin cos sin sin cos, , , , , ,2 2 2 22 1

58 2. Frame Fields

all s in I is called the curvature function of b. Thus k � 0, and the larger kis, the sharper the turning of b.

To carry this analysis further, we impose the restriction that k is never zeroso k > 0. The unit-vector field N = T ¢/k on b then tells the direction in which bis turning at each point. N is called the principal normal vector field of b (Fig.2.7). The vector field B = T ¥ N on b is called the binormal vector field of b.

3.1 Lemma Let b be a unit-speed curve in R3 with k > 0. Then the threevector fields T, N, and B on b are unit vector fields that are mutually orthog-onal at each point. We call T, N, B the Frenet frame field on b.

Proof. By definition � T � = 1. Since k = � T ¢ � > 0,

We saw above that T and N are orthogonal—that is, T • N = 0. Then byapplying Lemma 1.8 at each point, we conclude that � B � = 1, and B isorthogonal to both T and N. ◆

In summary, we have T = b¢, N = T ¢/k, and B = T ¥ N, satisfying T • T =N • N = B • B = 1, with all other dot products zero.

The key to the successful study of the geometry of a curve b is to use itsFrenet frame field T, N, B whenever possible, instead of the natural framefield U1, U2, U3. The Frenet frame field of b is full of information about b,whereas the natural frame field contains none at all.

The first and most important use of this idea is to express the derivativesT ¢, N¢, B¢ in terms of T, N, B. Since T = b¢, we have T ¢ = b≤ = kN. Nextconsider B¢. We claim that B¢ is, at each point, a scalar multiple of N. Toprove this, it suffices by orthonormal expansion to show that B¢ • B = 0 andB¢ • T = 0. The former holds since B is a unit vector. To prove the latter, dif-ferentiate B • T = 0, obtaining B¢ • T + B • T ¢ = 0; then

¢ = - ¢ = - =B T B T B N• • • .k 0

N T= ( ) ¢ =1 1k .

2.3 The Frenet Formulas 59

FIG. 2.7

Thus we can now define the torsion function t of the curve b to be the real-valued function on the interval I such that B¢ = -tN. (The minus sign is tra-ditional.) By contrast with curvature, there is no restriction on the values oft—it may be positive, negative, or zero at various points of I. We shallpresently show that t does measure the torsion, or twisting, of the curve b.

3.2 Theorem (Frenet formulas). If b: I Æ R3 is a unit-speed curve withcurvature k > 0 and torsion t, then

Proof. As we saw above, the first and third formulas are essentially justthe definitions of curvature and torsion. To prove the second, we use ortho-normal expansion to express N¢ in terms of T, N, B:

These coefficients are easily found. Differentiating N • T = 0, we get N¢ • T + N • T ¢ = 0; hence

As usual, N¢ • N = 0, since N is a unit vector field. Finally,

◆

3.3 Example We compute the Frenet frame T, N, B and the curvatureand torsion functions of the unit-speed helix

where c = (a2 + b2)1/2 and a > 0. Now

Hence

¢( ) = - -ÊË

ˆ¯T s

ac

sc

ac

sc2 2

0cos sin ., ,

T s sac

sc

ac

sc

bc

( ) = ¢( ) = -ÊË

ˆ¯b sin cos ., ,

b s asc

asc

bsc

( ) = ÊË

ˆ¯cos sin, , ,

¢ = - ¢ = - -( ) =N B N B N N• • • .t t

¢ = - ¢ = - = -N T N T N N• • • .k k

¢ = ¢ + ¢ + ¢N N T T N N N N B B• • • .

¢ =¢ = - +¢ = -

T N

N T B

B N

kk t

t

,

,

.

60 2. Frame Fields

Thus

Since T ¢ = kN, we get

Note that regardless of what values a and b have, N always points straight intoward the axis of the cylinder on which b lies (Fig. 2.8).

Applying the definition of cross product to B = T ¥ N gives

It remains to compute torsion. Now,

and by definition, B¢ = -tN. Comparing the formulas for B¢ and N, we con-clude that

So the torsion of the helix is also constant.Note that when the parameter b is zero, the helix reduces to a circle of

radius a. The curvature of this circle is k = 1/a (so the smaller the radius, thelarger the curvature), and the torsion is identically zero.

This example is a very special one—in general (as the examples in the exer-cises show) neither the curvature nor the torsion functions of a curve needbe constant.

t sbc

ba b

( ) = =+2 2 2

.

¢( ) = ÊË

ˆ¯B s

bc

sc

bc

sc2 2

0cos sin, , ,

B sbc

sc

bc

sc

ac

( ) = -ÊË

ˆ¯sin cos ., ,

N ssc

sc

( ) = - -ÊË

ˆ¯cos sin ., , 0

k s T sac

aa b

( ) = ¢( ) = =+

>2 2 2

0.


FIG. 2.8

3.4 Remark We have emphasized all along the distinction between atangent vector and a point of R3. However, Euclidean space has, as we haveseen, the remarkable property that given a point p, there is a natural one-to-one correspondence between points (v1, v2, v3) and tangent vectors (v1, v2, v3)p

at p. Thus one can transform points into tangent vectors (and vice versa) bymeans of this canonical isomorphism. In the next two sections particularly,it will often be convenient to switch quietly from one to the other withoutchange of notation. Since corresponding objects have the same Euclidean coor-dinates, this switching can have no effect on scalar multiplication, addition,dot products, differentiation, or any other operation defined in terms ofEuclidean coordinates.

Thus a vector field Y = ( y1, y2, y3)b on a curve b becomes itself a curve ( y1,y2, y3) in R3. In particular, if Y is parallel, its Euclidean coordinate functionsare constant, so Y is identified with a single point of R3.

A plane in R3 can be described as the union of all the perpendiculars to a given line at a given point. In vector language then, the plane through porthogonal to q π 0 consists of all points r in R3 such that (r - p) • q = 0. Bythe remark above, we may picture q as a tangent vector at p as shown in Fig. 2.9.

We can now give an informative approximation of a given curve near anarbitrary point on the curve. The goal is to show how curvature and torsioninfluence the shape of the curve. To derive this approximation we use a Taylorapproximation of the curve—and express this in terms of the Frenet frameat the selected point.

For simplicity, we shall consider the unit-speed curve b = (b1, b2, b3) nearthe point b(0). For s small, each coordinate bi(s) is closely approximated bythe initial terms of its Taylor series:

b bb b b

i ii i is

dds

sdds

s dds

s( ) ( ) + ( ) + ( ) + ( )~ .0 0 02

06

2

2

2 3

3

3

62 2. Frame Fields

FIG. 2.9

Hence

But b¢(0) = T0, and b≤(0) = k0N0, where the subscript indicates evaluation ats = 0, and we assume k0 π 0. Now

Thus by the Frenet formula for N¢, we get

Finally, substitute these derivatives into the approximation of b(s) givenabove, and keep only the dominant term in each component (that is, the onecontaining the smallest power of s). The result is

Denoting the right side by b(s), we obtain a curve b called the Frenet approxi-mation of b near s = 0. We emphasize that b has a different Frenet approx-imation near each of its points; if 0 is replaced by an arbitrary number s0,then s is replaced by s - s0, as usual in Taylor expansions.

Let us now examine the Frenet approximation given above. The first termin the expression for b is just the point b(0). The first two terms give thetangent line s Æ b(0) + sT0 of b at b(0)—the best linear approximation of bnear b(0). The first three terms give the parabola

which is the best quadratic approximation of b near b(0). Note that thisparabola lies in the plane through b(0) orthogonal to B0, the osculating planeof b at b(0). This parabola has the same shape as the parabola y = k0x2/2 inthe xy plane, and is completely determined by the curvature k0 of b at s = 0.

Finally, the torsion t0, which appears in the last and smallest term of b ,controls the motion of b orthogonal to its osculating plane at b(0), as shownin Fig. 2.10.

On the basis of this discussion, it is a reasonable guess that if a unit-speedcurve has curvature identically zero, then it is a straight line. In fact, this followsimmediately from (2) of Lemma 2.3, since k = � T ¢ � = � b≤ �, so that k = 0 if and only if b≤ = 0. Thus curvature does measure deviation from straightness.

s sT s NÆ ( ) + + ( )b k0 20 02

0 ,

b b k k ts sTs

Ns

B( ) ( ) + + +~ .02 60 0

2

0 0 0

3

0

¢¢¢( ) = - + ( ) +b kk

k t0 002

0 0 0 0 0Tdds

N B .

¢¢¢ = ( )¢ = + ¢b kk

kNdds

N N .

b b b b bs ss s( ) ( ) + ¢( ) + ¢¢( ) + ¢¢¢( )~ .0 02

06

02 3


A plane curve in R3 is a curve that lies in a single plane of R3. Evidently aplane curve does not twist in as interesting a way as even the simple helix inExample 3.3. The discussion above shows that for s small the curve b tendsto stay in its osculating plane at b(0); it is t0 π 0 that causes b to twist out ofthe osculating plane. Thus if the torsion of b is identically zero, we may wellsuspect that b never leaves this plane.

3.5 Corollary Let b be a unit-speed curve in R3 with k > 0. Then b is aplane curve if and only if t = 0.

Proof. Suppose b is a plane curve. Then by the remarks above, there existpoints p and q such that (b(s) - p) • q = 0 for all s. Differentiation yields

Thus q is always orthogonal to T = b¢ and N = b≤/k. But B is also orthog-onal to T and N, so, since B has unit length, B = ±q/�q�. Thus B¢ = 0, andby definition t = 0 (Fig. 2.11).

Conversely, suppose t = 0. Thus B¢ = 0; that is, B is parallel and maythus be identified (by Remark 3.4) with a point of R3. We assert that b liesin the plane through b(0) orthogonal to B. To prove this, consider the real-valued function

f s s B s( ) = ( ) - ( )( )b b 0 • .for all

¢( ) = ¢¢( ) =b bs s s• • .q q 0 for all

64 2. Frame Fields

FIG. 2.10

Then

But obviously, f(0) = 0, so f is identically zero. Thus

which shows that b lies entirely in this plane orthogonal to the (parallel)binormal of b. ◆

We saw at the end of Example 3.3 that a circle of radius a has curvature1/a and torsion zero. Furthermore, the formula given there for the principalnormal shows that for a circle, N always points toward its center. This sug-gests how to prove the following converse.

3.6 Lemma If b is a unit-speed curve with constant curvature k > 0 andtorsion zero, then b is part of a circle of radius 1/k.

Proof. Since t = 0, b is a plane curve. What we must now show is thatevery point of b is at distance 1/k from some fixed point—which will thusbe the center of the circle. Consider the curve g = b + (1/k)N. Using thehypothesis on b, and (as usual) a Frenet formula, we find

Hence the curve g is constant; that is, b(s) + (1/k)N(s) has the same value,say c, for all s (see Fig. 2.12). But the distance from c to b(s) is

◆d s s N sc c, b bk k

( )( ) = - ( ) = ( ) =1 1

.

¢ = ¢ + ¢ = + -( ) =g bk k

k1 1

0N T T .

b bs B s( ) - ( )( )0 • for all ,

dfds

B T B= ¢ = =b • • .0


FIG. 2.11 FIG. 2.12

In principle, every geometric problem about curves can be solved by meansof the Frenet formulas. In simple cases it may be just enough to record thedata of the problem in convenient form, differentiate, and use the Frenet for-mulas. For example, suppose b is a unit-speed curve that lies entirely in thesphere of radius a centered at the origin of R3. To stay in the sphere, bmust curve; in fact it is a reasonable guess that the minimum possible curva-ture occurs when b is on a great circle of . Such a circle has radius a, sowe conjecture that a spherical curve b has curvature k � 1/a, where a is theradius of its sphere.

To prove this, observe that since every point of has distance a from theorigin, we have b • b = a2. Differentiation yields 2b¢ • b = 0, that is, b • T =0. Another differentiation gives b¢ • T + b • T ¢ = 0, and by using a Frenetformula we get T • T + kb • N = 0; hence

By the Schwarz inequality,

and since k � 0 we obtain the required result:

Continuation of this procedure leads to a necessary and sufficient condition(expressed in terms of curvature and torsion) for a curve to be spherical, thatis, lie on some sphere in R3 (Exercise 10).

Exercises

1. Compute the Frenet apparatus k, t, T, N, B of the unit-speed curve b(s) = (4/5 coss, 1 - sins, -3/5 coss). Show that this curve is a circle; find itscenter and radius.

2. Consider the curve

defined on I: -1 < s < 1. Show that b has unit speed, and compute its Frenetapparatus.

b ss s s( ) =

+( ) -( )ÊËÁ

ˆ¯

13

13 2

3 2 3 2

, ,

k kb

= =1 1•

.N a

�

b b• N N a� = ,

kb • .N = -1

Â

Â

Â

66 2. Frame Fields

3. For the helix in Example 3.3, check the Frenet formulas by direct sub-stitution of the computed values of k, t, T, N, B.

4. Prove that

(A formal proof uses properties of the cross product established in the Exer-cises of Section 1—but one can recall these formulas by using the right-handrule given at the end of that section.)

5. If A is the vector field tT + kB on a unit-speed curve b, show that theFrenet formulas become

6. A unit-speed parametrization of a circle may be written

where ei • ej = dij.If b is a unit-speed curve with k(0) > 0, prove that there is one and only

one circle g that approximates b near b(0) in the sense that

Show that g lies in the osculating plane of b at b(0) and find its center c andradius r (see Fig. 2.13). The circle g is called the osculating circle and c thecenter of curvature of b at b(0). (The same results hold when 0 is replaced byany number s.)

g b g b g b0 0 0 0 0 0( ) = ( ) ¢( ) = ¢( ) ¢¢( ) = ¢¢( ), , and .

g s rsr

rsr

( ) = + +c e ecos sin1 2,

¢ = ¥

¢ = ¥

¢ = ¥

T A T

N A N

B A B

,

,

.

T N B B N

N B T T B

B T N N T

.

= ¥ = - ¥

= ¥ = - ¥

= ¥ = - ¥

,

,


FIG. 2.13

7. If a and a reparametrization = a(h) are both unit-speed curves, showthat

(a) h(s) = ± s + s0 for some number s0;(b) = ±T(h),

= N(h), = k (h), t = t (h),= ±B(h),

where the sign (±) is the same as that in (a), and we assume k > 0. Thus evenin the orientation-reversing case, the principal normals N and still pointin the same direction.

8. Curves in the plane. For a unit-speed curve b(s) = (x(s), y(s)) in R2, theunit tangent is T = b¢ = (x¢, y¢) as usual, but the unit normal N is defined byrotating T through +90°, so N = (-y¢, x¢). Thus T ¢ and N are collinear, andthe plane curvature k of b is defined by the Frenet equation T ¢ = kN.

(a) Prove that k = T ¢ • N and N¢ = -kT.(b) The slope angle j(s) of b is the differentiable function such that

(The existence of j derives from Ex. 12 of Sec. 1.) Show that k = j¢.(c) Find the curvature k of the following plane curves.

(i) (rcos , rsin ), counterclockwise circle.

(ii) (rcos(- ), rsin(- )), clockwise circle.

(d) Show that if k does not change sign, then |k | is the usual R3 curvaturek. (For such comparisons we can always regard R2 as, say, the xy plane in R3.)

9. Let b be the Frenet approximation of a unit-speed curve b with t π 0near s = 0.

If, say, the B0 component of b is removed, the resulting curve is the orthog-onal projection of b in the T0N0 plane. It is the view of b ª b that one getsby looking toward b(0) = b (0) directly along the vector B0.

Sketch the general shape of the orthogonal projections of b near s = 0 ineach of the planes T0N0 (osculating plane), T0B0 (rectifying plane), and N0B0

(normal plane). These views of b ª b can be confirmed experimentally usinga bent piece of wire. For computer views, see Exercise 15 of Section 4.

10. Spherical curves. Let a be a unit-speed curve with k > 0, t π 0.(a) If a lies on a sphere of center c and radius r, show that

a r r s- = - - ¢c N B,

tr

tr

tr

tr

T U Ux y= ( ) = +cos sin cos sin .j j j j,

N

BtkN

T

a

68 2. Frame Fields

where r = 1/k and s = 1/t . Thus r2 = r2 + (r¢s)2.

(b) Conversely, if r2 + (r¢s)2 has constant value r2 and r¢ π 0, show thata lies on a sphere of radius r.(Hint: For (b), show that the “center curve” g = a + rN + r¢sB—suggestedby (a)—is constant.)

11. Let b, : I Æ R3 be unit-speed curves with nonvanishing curvature andtorsion. If T = , then b and are parallel (Ex. 10 of Sec. 2). IfB = , prove that is parallel to either b or the curve s Æ -b(s).

2.4 Arbitrary-Speed Curves

It is a simple matter to adapt the results of the previous section to the studyof a regular curve a: I Æ R3 that does not necessarily have unit speed. Wemerely transfer to a the Frenet apparatus of a unit-speed reparametrization

of a. Explicitly, if s is an arc length function for a as in Theorem 2.1, then

or, in functional notation, a = (s), as suggested by Fig. 2.14. Now if> 0, t , , , and are defined for as in Section 3, we define for a the

curvature function: k = (s),torsion function: t = (s),unit tangent vector field: T = (s),principal normal vector field: N = (s),binormal vector field: B = (s).

In general k and are different functions, defined on different intervals.But they give exactly the same description of the turning of the common routeof a and , since at any point a(t) = (s(t)) the numbers k(t) and (s(t)) arekaa

k

BN

Tt

k

aBNTtka

a at s t t( ) = ( )( ) for all ,

a

bBbT

b

2.4 Arbitrary-Speed Curves 69

FIG. 2.14

by definition the same. Similarly with the rest of the Frenet apparatus; sinceonly a change of parametrization is involved, its fundamental geometricmeaning is the same as before. In particular, T, N, B is again a frame field on a linked to the shape of a as indicated in the discussion of Frenet approximations.

For purely theoretical work, this simple transference is often all that isneeded. Data about a converts into data about the unit-speed reparame-trization ; results about convert to results about a. For example, if a isa regular curve with t = 0, then by the definition above has = 0; by Corollary 3.5, is a plane curve, so obviously a is too.

However, for explicit numerical computations—and occasionally for thetheory as well—this transference is impractical, since it is rarely possible tofind explicit formulas for . (For example, try to find a unit-speed parame-trization for the curve a(t) = (t, t2, t3).)

The Frenet formulas are valid only for unit-speed curves; they tell the rateof change of the frame field T, N, B with respect to arc length. However, thespeed v of the curve is the proper correction factor in the general case.

4.1 Lemma If a is a regular curve in R3 with k > 0, then

Proof. Let be a unit-speed reparametrization of a. Then by definition,T = (s), where s is an arc length function for a. The chain rule as appliedto differentiation of vector fields (Exercise 7 of Section 2) gives

By the usual Frenet equations, . Substituting the function s in thisequation yields

by the definition of k and N in the arbitrary-speed case. Since ds/dt is thespeed function v of a, these two equations combine to yield T¢ = kvN. Theformulas for N¢ and B¢ are derived in the same way. ◆

There is a commonly used notation for the calculus that completely ignoreschange of parametrization. For example, the same letter would designateboth a curve a and its unit-speed parametrization , and similarly with thea

¢( ) = ( ) ( ) =T s s N s Nk k

¢ =T Nk

¢ = ¢( )T T sdsdt

.

Ta

¢ =¢ = - +¢ = -

T vN

N vT vB

B vN

kk t

t

,

,

.

a

ata

aa

70 2. Frame Fields

Frenet apparatus of these two curves. Differences in derivatives are handledby writing, say, dT/dt for T ¢, but dT/ds for either or its reparametrization

(s). If these conventions were used, the proof above would combine thechain rule dT/dt = (dT/ds) (ds/dt) and the Frenet formula dT/ds = kN to givedT/dt = kvN.

Only for a constant-speed curve is acceleration always orthogonal to veloc-ity, since b¢ • b¢ constant is equivalent to (b¢ • b¢)¢ = 2b¢ • b≤ = 0. In the generalcase, we analyze velocity and acceleration by expressing them in terms of theFrenet frame field.

4.2 Lemma If a is a regular curve with speed function v, then the veloc-ity and acceleration of a are given by (Fig. 2.15.)

Proof. Since a = (s), where s is the arc length function of a, we find,using Lemma 4.5 of Chapter 1, that

Then a second differentiation yields

where we use Lemma 4.1. ◆

The formula a ¢ = vT is to be expected since a ¢ and T are each tangent to the curve and T has a unit length, while � a ¢ � = v. The formula for acceleration is more interesting. By definition, a≤ is the rate of change of the

¢¢ = + ¢ = +a kdvdt

T vTdvdt

T v N2 ,

¢ = ¢( ) = ( ) =a a sdsdt

vT s vT.

a

¢ = ¢¢ = +a a kvTdvdt

T v N, 2 .

¢T¢T


FIG. 2.15

velocity a ¢, and in general both the length and the direction of a ¢ are chang-ing. The tangential component (dv/dt)T of a≤ measures the rate of change ofthe length of a ¢ (that is, of the speed of a). The normal component kv2N mea-sures the rate of change of the direction of a ¢. Newton’s laws of motion showthat these components may be experienced as forces. For example, in a carthat is speeding up or slowing down on a straight road, the only force onefeels is due to (dv/dt)T. If one takes an unbanked curve at speed v, the result-ing sideways force is due to kv2N. Here k measures how sharply the road turns;the effect of speed is given by v2, so 60 miles per hour is four times as unset-tling as 30.

We now find effectively computable expressions for the Frenet apparatus.

4.3 Theorem Let a be a regular curve in R3. Then

Proof. Since v = � a ¢ � > 0, the formula T = a ¢/�a ¢� is equivalent to a ¢ = vT. From the preceding lemma we get

since T ¥ T = 0. Taking norms we find

because � B � = 1, k � 0, and v > 0. Indeed, this equation shows that forregular curves, � a ¢ ¥ a≤ � > 0 is equivalent to the usual condition k > 0.(Thus for k > 0, a ¢ and a≤ are linearly independent and determine the oscu-lating plane at each point, as do T and N.) Then

Since N = B ¥ T is true for any Frenet frame field (Exercise 4 of Section3), only the formula for torsion remains to be proved.

To find the dot product (a ¢ ¥ a≤) • a� we express everything in terms ofT, N, B. We already know that a ¢ ¥ a≤ = kv3B. Thus, since 0 = T • B =N • B, we need only find the B component of a�. But

Bv

=¢ ¥ ¢¢

=¢ ¥ ¢¢¢ ¥ ¢¢

a ak

a aa a3

.

¢ ¥ ¢¢ = =a a k kv B v3 3

¢ ¥ ¢¢ = ( ) ¥ +ÊË

ˆ¯

= ¥ + ¥ =

a a k

k k

vTdvdt

T v N

vdvdt

T T v T N v B

2

3 3 ,

T

N B T

B

= ¢ ¢

= ¥ = ¢ ¥ ¢¢ ¢= ¢ ¥ ¢¢ ¢ ¥ ¢¢ = ¢ ¥ ¢¢( ) ¢¢¢ ¢ ¥ ¢¢

a a

k a a aa a a a t a a a a a

,

, ,

,

3

2• .

72 2. Frame Fields

where we use Lemma 4.1. Consequently, (a ¢ ¥ a≤) • a � = k2v6t, and since � a ¢ ¥ a≤ � = kv3, we have the required formula for t. ◆

The triple scalar product in this formula for t could (by Exercise 4 ofSection 1) also be written a ¢ • a≤ ¥ a�. But we need a ¢ ¥ a≤ anyway, so it ismore efficient to find (a ¢ ¥ a≤) • a�.

4.4 Example We compute the Frenet apparatus of the 3-curve

The derivatives are

Now,

so

Applying the definition of cross product yields

Dotting this vector with itself, we get

Hence

The expressions above for a ¢ ¥ a≤ and a� yield

¢ ¥ ¢¢( ) ¢¢¢ =a a a• • • .6 18 2

¢( ) ¥ ¢¢( ) = +( )a at t t18 2 1 2 .

18 1 4 1 2 18 12 2 2 2 2 2 2 2 2( ) - +( ) + + +( )[ ] = ( ) +( )t t t t .

¢( ) ¥ ¢¢( ) = - +-

= - + - +( )a at t

U U U

t t t

t t

t t t181 2 1

1

18 1 2 11 2 3

2 2 2 2, , .

v t t t( ) = ¢( ) = +( )a 18 1 2 .

¢( ) ¢( ) = + +( )a at t t t• 18 1 2 2 4 ,

¢( ) = - +( )¢¢( ) = -( )¢¢¢( ) = -( )

a

a

a

t t t t

t t t

t

3 1 2 1

6 1

6 1 0 1

2 2, , ,

, , ,

, , .

a t t t t t t( ) = - +( )3 3 33 2 3, , .

¢¢¢ = +ÊË

ˆ¯

¢= ¢ +

= +

a k k

k t

dvdt

T v N v N

v B

2 2

3

. . .

. . . ,


It remains only to substitute this data into the formulas in Theorem 4.3, withN being computed by another cross product. The final results are

Alternatively, we could use the identity in Lemma 1.8 to compute � a ¢ ¥a≤ � and express

as a determinant by Exercise 4 of Section 1.To summarize, we now have the Frenet apparatus for an arbitrary regular

curve a, namely, its curvature, torsion, and Frenet frame field. This appara-tus satisfies the extended Frenet formulas with speed factor v and can be com-puted by Theorem 4.3. If v = 1, that is, if a is a unit-speed curve, the resultsof Section 3 are recovered.

Let us consider some applications of the Frenet formulas. There are anumber of natural ways in which a given curve b gives rise to a new curve b whose geometric properties illuminate some aspect of the behavior of b.

For example, the spherical image of a unit-speed curve b is the curve s ª T with the same Euclidean coordinates as T = b¢. Geometrically, s isgotten by moving each T(s) to the origin of R3, as suggested in Fig. 2.16.Thus s lies on the unit sphere S, and the motion of s represents the turningof b.

¢ ¥ ¢¢( ) ¢¢¢ = ¢ ¢¢ ¥ ¢¢¢( )a a a a a a• •

Tt t t

t

Nt t

t

Bt t t

t

t

=- +( )

+( )

=- -( )

+

=- + - +( )

+( )

= =+( )

1 2 1

2 1

2 1 01

1 2 1

2 1

1

3 1

2 2

2

2

2

2 2

2

2 2

, ,,

, ,,

, ,,

k t .

74 2. Frame Fields

FIG. 2.16

For instance, if b is the helix in Example 3.3, the formula there for T showsthat

So the spherical image of a helix lies on the circle cut from by the plane z = b/c.

Although the original curve b has unit speed, we cannot expect that sdoes also. In fact, s = T implies s ¢ = T ¢ = kN, so the speed of s equals thecurvature k of b. Thus to compute the curvature of s, we must use theextended Frenet formulas in Theorem 4.3. From

we get

By Theorem 4.3 the curvature of the spherical image s is

and thus depends only on the ratio of torsion to curvature for the originalcurve b.

Here is a closely related application in which this ratio t/k turns out to bedecisive.

4.5 Definition A regular curve a in R3 is a cylindrical helix provided theunit tangent T of a has constant angle J with some fixed unit vector u; thatis, T(t) • u = cosJ for all t.

This condition is not altered by reparametrization, so for theoretical pur-poses we need only deal with a cylindrical helix b that has unit speed. Sosuppose b is a unit-speed curve with T • u = cosJ. If we pick a referencepoint, say b(0), on b, then the real-valued function

tells how far b(s) has “risen” in the u direction since leaving b(0) (Fig. 2.17).But

h s s( ) = ( ) - ( )( )b b 0 • u

ks s k t

ktks =

¢ ¥ ¢¢=

+= + Ê

Ëˆ¯

ÊËÁ

ˆ¯

≥v3

2 2 2 1 2

1 1

¢ ¥ ¢¢ = - ¥ + ¥ = +( )s s k k t k k t3 2 2N T N B B T .

¢¢ = ( )¢ = + ¢ = - + +s kk

k kk

ktNdds

N N Tdds

N B2 ,

Â

s sac

sc

ac

sc

bc

( ) = -ÊË

ˆ¯sin cos ., ,


so b is rising at a constant rate relative to arc length, and h(s) = s cosJ. If weshift to an arbitrary parametrization, this formula becomes

where s is the arc length function.By drawing a line through each point of b in the u direction, we construct

a cylinder C on which b moves in such a way as to cut each such line at con-stant angle J, as in Fig. 2.18. In the special case when this cylinder is circu-lar, b is evidently a helix of the type defined in Example 3.3.

It turns out to be quite easy to identify cylindrical helices.

4.6 Theorem A regular curve a with k > 0 is a cylindrical helix if andonly if the ratio t/k is constant.

Proof. It suffices to consider the case where a has unit speed. If a is acylindrical helix with T • u = cosJ, then

Since k > 0, we conclude that N • u = 0. Thus for each s, u lies in the planedetermined by T(s) and B(s). Orthonormal expansion yields

u = +cos sin .J JT B

0 = ( )¢ = ¢ =T T N• • • .u u uk

h t s t( ) = ( ) cos J,

dhds

T= ¢ = =b J• • cosu u ,

76 2. Frame Fields

FIG. 2.17 FIG. 2.18

As usual we differentiate and apply Frenet formulas to obtain

Hence t sinJ = kcosJ, so that t/k has constant value cotJ.Conversely, suppose that t/k is constant. Choose an angle J such that

cotJ = t/k. If

we find

This parallel vector field U then determines (as in Remark 3.4) a unit vectoru such that T • u = cosJ, so a is a cylindrical helix. ◆

In Exercise 9 this information about cylindrical helices is used to show thatcircular helices are characterized by constancy of curvature and torsion (seealso Corollary 5.5 of Chapter 3).

Simple hypotheses on a regular curve in R3 thus have the following effects(¤ means “if and only if”):

Exercises

Computer commands that produce the Frenet apparatus, k, t, T, N, B, of acurve are given in the Appendix. Their use is optional in the following exercises.

1. For the curve a(t) = (2t, t2, t3/3),(a) Compute the Frenet apparatus.(b) Sketch the curve for -4 � t � 4, showing T, N, B at t = 2.(c) Find the limiting values of T, N, and B as t Æ -• and t Æ •.

2. Express the curvature and torsion of the curve a(t) = (cosh t, sinh t, t)in terms of arc length s measured from t = 0.

ktk tk tt k

= ¤= ¤

> = ¤> > ¤

π ¤

0

0

0 0

0 0

0

straight line,

plane curve,

const and circle,

const and const circular helix,

const cylindrical helix.

¢ = -( ) =U Nk J t Jcos sin .0

U T B= +cos sinJ J ,

0 = -( )k J t Jcos sin .N


3. The curve a(t) = (tcos t, tsin t, t) lies on a double cone and passesthrough the vertex at t = 0.

(a) Find the Frenet apparatus of a at t = 0.(b) Sketch the curve for -2p � t � 2p, showing T, N, B at t = 0.

4. Show that the curvature of a regular curve in R3 is given by

5. If a is a curve with constant speed c > 0, show that

where for N, B, t, we assume a≤ never zero, that is, k > 0.

6. (a) If a is a cylindrical helix, prove that its unit vector u (Thm. 4.5) is

and the coefficients here are cosJ and sinJ (for J as in Def. 4.5).(b) Check (a) for the cylindrical helix in Example 4.2 of Chapter 1.

7. Let a: I Æ R3 be a cylindrical helix with unit vector u. For t0 Œ I, thecurve

is called a cross-sectional curve of the cylinder on which a lies. Prove:(a) g lies in the plane through a(t0) orthogonal to u.(b) The curvature of g is k/sin2 J, where k is the curvature of a.

8. Verify that the following curves are cylindrical helices and, for each, findthe unit vector u, angle J, and cross-sectional curve s.

(a) The curve in Exercise 1. (b) The curve in Example 4.4.(c) The curve in Exercise 2.

9. If a is a curve with k > 0 and t both constant, show that a is a circularhelix.

10. (a) Prove that a curve is a cylindrical helix if and only if its sphericalimage is part of a circle.

(b) Sketch the spherical image of the cylindrical helix in Exercise 1. Is ita complete circle? Find its center.

g a a at t t t( ) = ( ) - ( ) - ( )( )( )0 • u u

u =+

++

tk t

kk t2 2 2 2

T B,

T c N B c

c c

= ¢ = ¢¢ ¢¢ = ¢ ¥ ¢¢ ¢¢( )

=¢¢

=¢ ¥ ¢¢ ¢¢¢

¢¢

a a a a a a

ka

ta a a

a

, , ,

, ,2 2 2

•

k a2 4 2 2v dv dt= ¢¢ - ( ) .

78 2. Frame Fields

11. If a is a curve with k > 0, its central curve a* = a + (1/k)N consists ofall centers of curvature of a (Ex. 6 of Sec. 3). For nonzero numbers a andb, let bab be the helix in Example 3.3.

(a) Show that the central curve of bab is the helix bâb, where â = -b2/a.(b) Deduce that the central curve of bâb is the original helix bab.(c) (Computer graphics.) Plot three complete turns of the mutually central

helices b2,1 and b-1/2,1 in the same figure.

12. If a(t) = (x(t), y(t)) is a regular curve in R2, show that its plane curva-ture (Ex. 8 of Sec. 3) is given by

where J is the rotation operator J(a, b) = (-b, a).

13. (Continuation.) For a plane curve a with k π 0, the central curve a* = a + (1/k )N is called the evolute of a. Thus a* gives a direct pointwisedescription of the turning of a.

(a) Show that

(b) Find a formula for the line segment lt from a(t) to a*(t). This segmentis the radius (line) of the approximating circle to a near a(t) (Ex. 6 ofSec. 3)(c) Prove that lt is normal to a at a(t) and tangent to a* at a*(t). (Hint:It can be assumed that a has unit speed.)

14. (Continuation, Computer graphics.) In each case, plot the given planecurve and its evolute on the same figure, showing some of the constructionlines lt.

(a) The ellipse a(t) = (2cos t, sin t).(b) The cycloid a(t) = (t + sin t, 1 + cos t) for -2p £ t £ 2p. (Here the evolutebears an unexpected relation to the original curve.)

15. (Computer continuation of Ex. 9 of Sec. 3.)(a) Write the commands that, given a regular curve a with k(0) > 0, plot—on a small interval -e � t � e —the orthogonal projection of a into theosculating, rectifying, and normal planes at a(0). Show the projections ascurves in R2.(b) Test (a) on the curves (3), (4), (5) in Example 4.2 of Chapter 1 andthose in Example 4.3 of Chapter 3. Compare results.

a aa a

a aa* = +

¢ ¢¢¢ ¢( ) ¢( )•

•.

JJ

˜•

ka a

=¢¢ ¢( )

=¢ ¢¢ - ¢¢ ¢¢ + ¢( )

Jv

x y x y

x y3 2 2 3 2 ,


The following exercise shows that the condition k > 0 cannot beavoided in a detailed study of the geometry of curves in R3 for even ifk is zero at only a single point, the geometric character of the curvecan change radically at that point. (This difficulty does not arise forcurves in the plane.)

16. It is shown in advanced calculus that the function

is infinitely differentiable (has continuous derivatives of all orders). Thus

is a well-defined differentiable curve.(a) Sketch a on an interval -a � t � a.(b) Show that the curvature of a is zero only at t = 0.(c) What are the osculating planes of a for t < 0 and t > 0?

In the following exercise, a global geometric invariant of curves is gotten byintegrating a local invariant.

17. The total curvature of a unit-speed curve a: I Æ R3 is . If a is

merely regular, the formula becomes . Find the total curvature of

the following curves:(a) The curve in Example 4.4.(b) The helix in Example 3.3.(c) The curve in Exercise 2.(d) The ellipse a(t) = (acos t, bsin t) on 0 � t � 2p.

18. One definition of convexity for a smoothly closed plane curve is thatits curvature k is positive (hence its plane curvature k is either always posi-tive or always negative). Prove that a convex closed plane curve has total cur-vature 2p. (Hint: Consider its spherical image.)

A theorem of Fenchel asserts that every regular closed curve a in R3 hastotal curvature �2p. Surprisingly, this has an easy proof in terms of surfacetheory (see Sec. 8 of Ch. 6).

19. (Computer.)(a) Plot the curve

Even looking at this curve from different viewpoints may not make its cross-ing pattern clear, but Exercise 21 of Section 5.4 will show that t is a trefoil knot.

t pt t t t t t t( ) = + -( )4 2 2 4 2 2 3 0 2cos cos sin sin sin ., , on � �

k t v t dtI

( ) ( )Úk s ds

I( )Ú

a t t f t f t( ) = ( ) -( )( ), ,

f tt

e tt( ) =

£>

ÏÌÓ -

0 0

01 2

if ,

if .

80 2. Frame Fields

(Intuitively, a simple closed curve in R3 is a knot provided it cannot be contin-uously deformed—always remaining simply closed—until it becomes a circle.)

The Fary-Milnor theorem asserts that every knot has total curvaturestrictly greater than 4p. Show:

(b) The plane curve obtained from t by removing the z-component sin3thas total curvature exactly 4p. (This curve is not simply closed, and henceis not a knot.)(c) t can be deformed to a knot that has (numerically estimated) total curvature less than 4.01p.

20. (Computer.)(a) Write a command that, given an arbitrary regular curve, returns thetest function in Exercise 10 of Section 3 whose constancy implies that thecurve lies on a sphere. (Plotting this function provides a good test for con-stancy and does not require simplifying it.) (Hint: To allow for arbitraryparametrization, replace derivatives f ¢(s) by f ¢(t)v(t), where v(t) = ds/dt.)(b) In each case, decide whether the curve lies on a sphere, and if so, findits radius and center:

(i) a(t) = (2sin t, sin2t, 2sin2 t);(ii) b(t) = (cos2 t, sin2t, 2sin t);

(iii) g (t) = (cos t, 1 + sin t, 2sin ).

21. Prove that the cubic curve g (t) = (at, bt2, ct3), abc π 0, is a cylindricalhelix if and only if 3ac = ±2b2. (Computer optional.)

2.5 Covariant Derivatives

In Chapter 1 the definition of a new object (curve, differential form, map-ping, . . .) was usually followed by an appropriate notion of derivative of thatobject. To see how to define the derivative of a vector field on a Euclideanspace, we mimic the definition of the derivative v[ f ] of a function f relative toa tangent vector v at a point p (Definition 3.1 of Chapter 1). In fact, replacingf by a vector field W on R3 gives a vector field t Æ W(p + tv) on the curve t Æ p + tv. The derivative of such a vector field was defined in Section 2. Thenthe derivative of W with respect to v will be the derivative of t Æ W(p + tv) att = 0.

5.1 Definition Let W be a vector field on R3, and let v be a tangent vectorfield to R3 at the point p. Then the covariant derivative of W with respect tov is the tangent vector

at the point p.— = +( )¢( )vW W tp v 0

t2

2.5 Covariant Derivatives 81

Evidently —vW measures the initial rate of change of W(p) as p moves in thev direction. (The term “covariant” derives from the generalization of thisnotion discussed in Chapter 7.)

For example, suppose W = x2U1 + yzU3, and

Then

so

where strictly speaking U1 and U3 are also evaluated at p + tv. Thus,

5.2 Lemma If W = wiUi is a vector field on R3, and v is a tangentvector at p, then

Proof. We have

for the restriction of W to the curve t Æ p + tv. To differentiate such avector field (at t = 0), one simply differentiates its Euclidean coordinates(at t = 0). But by the definition of directional derivative (Definition 3.1 ofChapter 1), the derivative of wi(p + tv) at t = 0 is precisely v[wi]. Thus

◆

In short, to apply —v to a vector field, apply v to its Euclidean coordinates.Thus the following linearity and Leibnizian properties of covariant deriva-tive follow easily from the corresponding properties (Theorem 3.3 of Chapter1) of directional derivatives.

5.3 Theorem Let v and w be tangent vectors to R3 at p, and let Y and Zbe vector fields on R3. Then for numbers a, b and functions f,

(1) —av+bwY = a—vY + b—wY.(2) —v(aY + bZ) = a—vY + b—vZ.(3) —v(fY) = v[f]Y(p) + f(p)—vY.(4) v[Y • Z] = —vY • Z(p) + Y(p) • —vZ.

— = +( )¢( ) = [ ] ( )Âv i iW W t w Up v v p0 .

W t w t U ti ip v p v p v+( ) = +( ) +( )Â

— = [ ] ( )Âv i iW w Uv p .

Â

— = +( )¢( ) = - ( ) + ( )vW W t U Up v p p0 4 21 3 .

W t t U tUp v+( ) = -( ) +2 221 3,

p v+ = -( )t t t2 1 2, ,,

v p= -( ) = ( )1 0 2 2 1 0, , at , , .

82 2. Frame Fields

Proof. For example, let us prove (4). If

then

Hence by Theorem 3.3 of Chapter 1,

But by the preceding lemma,

Thus the two sums displayed above are precisely —vY • Z(p) and Y(p) • —vZ.◆

Using the pointwise principle (Chapter 1, Section 2), we can take thecovariant derivative of a vector field W with respect to a vector field V, ratherthan a single tangent vector v. The result is the vector field —VW whose valueat each point p is —V(p)W. Thus —VW consists of all the covariant derivativesof W with respect to the vectors of V. It follows immediately from the lemmaabove that if W = wiUi, then

Coordinate computations are easy using the basic identity Ui[ f ] = ∂f/∂xi.For example, suppose V = (y - x)U1 + xyU3 and (as in the example above)W = x2U1 + yzU3. Then

Hence

For the covariant derivative —VW as expressed entirely in terms ofvector fields, the properties in the preceding theorem take the following form.

5.4 Corollary Let V, W, Y, and Z be vector fields on R3. Then(1) —fV+gWY = f —VY + g—WY, for all functions f and g.(2) —V(aY + bZ) = a—VY + b—VZ, for all numbers a and b.

— = -( ) +VW x y x U xy U2 12

3.

V x y x U x x y x

V yz xyU yz xy

21

2

32

2[ ] = -( ) [ ] = -( )[ ] = [ ] =

,

.

— = [ ]ÂV i iW V w U .

Â

— = [ ] ( ) — = [ ] ( )Â Âv i i v i iY y U Z z Uv p v pand .

v v v p p vY Z y z y z y zi i i i i i• .[ ] = [ ] = [ ] ( ) + ( ) [ ]Â Â Â

Y Z y zi i• .= Â

Y yU Z zUi i i i= =Â Âand ,

2.5 Covariant Derivatives 83

(3) —V(fY) = V[ f ]Y + f —VY, for all functions f.(4) V [Y • Z] = —VY • Z + Y • —VZ.

We shall omit the proof, which is an exercise in the use of parentheses basedon the (pointwise principle) definition (—VY )(p) = —V(p)Y.

Note that —VY does not behave symmetrically with respect to V and Y. Thisis to be expected, since it is Y that is being differentiated, while the role of Vis merely algebraic. In particular, —fVY is f —VY, but —V( fY) is not f —VY: Thereis an extra term arising from the differentiation of f by V.

Exercises

1. Consider the tangent vector v = (1, -1, 2) at the point p = (1, 3, -1).Compute —vW directly from the definition, where

(a) W = x2U1 + yU2. (b) W = xU1 + x2U2 - z2U3.

2. Let V = -yU1 + xU3 and W = cosxU1 + sinxU2. Express the follow-ing covariant derivatives in terms of U1, U2, U3:

(a) —VW. (b) —VV.(c) —V (z2W ). (d) —W (V ).(e) —V (—vW ). (f) —V (xV - zW ).

3. If W is a vector field with constant length �W �, prove that for any vectorfield V, the covariant derivative —VW is everywhere orthogonal to W.

4. Let X be the special vector field xiUi, where x1, x2, x3 are the naturalcoordinate functions of R3. Prove that —VX = V for every vector field V.

5. Let W be a vector field defined on a region containing a regular curve a.Then t Æ W(a(t)) is a vector field on a called the restriction of W to a anddenoted by Wa.

(a) Prove that —a¢(t)W = (Wa)¢ (t).(b) Deduce that the straight line in Definition 5.1 may be replaced by anycurve with initial velocity v. Thus the derivative Y¢ of a vector field Y ona curve a is (almost) —a ¢Y.

2.6 Frame Fields

When the Frenet formulas were discovered (by Frenet in 1847, and indepen-dently by Serret in 1851), the theory of surfaces in R3 was already a richlydeveloped branch of geometry. The success of the Frenet approach to curves

Â

84 2. Frame Fields

led Darboux (around 1880) to adapt this “method of moving frames” to thestudy of surfaces. Then, as we mentioned earlier, it was Cartan who broughtthe method to full generality. His essential idea was very simple: To each pointof the object under study (a curve, a surface, Euclidean space itself, . . .)assign a frame; then using orthonormal expansion express the rate of changeof the frame in terms of the frame itself. This, of course, is just what theFrenet formulas do in the case of a curve.

In the next three sections we shall carry out this scheme for the Euclideanspace R3. We shall see that geometry of curves and surfaces in R3 is not merelyan analogue, but actually a corollary, of these basic results. Since the mainapplication (to surface theory) comes only in Chapter 6, these sections maybe postponed, and read later as a preliminary to that chapter.

By means of the pointwise principle (Chapter 1, Section 2) we can automatically extend operations on individual tangent vectors to operationson vector fields. For example, if V and W are vector fields on R3, then thedot product V • W of V and W is the (differentiable) real-valued function on R whose value at each point p is V(p) • W(p). The norm �V� of V is thereal-valued function on R3 whose value at p is �V(p)�. Thus �V � = (V • V )1/2.By contrast with V • W, the norm function �V � need not be differentiable atpoints for which V(p) = 0, since the square-root function is badly behaved at 0.

In Chapter 1 we called the three vector fields U1, U2, U3 the natural framefield on R3. Here is a simple but crucial generalization.

6.1 Definition Vector fields E1, E2, E3 on R3 constitute a frame field onR3 provided

where dij is the Kronecker delta.

Thus at each point p the vectors E1(p), E2(p), E3(p) do in fact form a frame(Definition 1.4) since they have unit length and are mutually orthogonal.

In elementary calculus, frame fields are usually derived from coordinatesystems, as in the following cases.

6.2 Example (1) The cylindrical frame field (Fig. 2.19). Let r, J, z be theusual cylindrical coordinate functions on R3. We shall pick a unit vector fieldin the direction in which each coordinate increases (when the other two areheld constant). For r, this is evidently

E U U1 1 2= +cos sinJ J ,

E E i ji j ij• = £ £( )d 1 3, ,

2.6 Frame Fields 85

pointing straight out from the z axis. Then

points in the direction of increasing J as in Fig. 2.19. Finally, the directionof increase of z is, of course, straight up, so

It is easy to check that Ei • Ej = dij, so this is a frame field (defined on allof R3 except the z axis). We call it the cylindrical frame field on R3.

(2) The spherical frame field on R3 (Fig. 2.20). In a similar way, a framefield F1, F2, F3 can be derived from the spherical coordinate functions r, J, jon R3. As indicated in the figure, we shall measure j up from the xy planerather than (as is usually done) down from the z axis.

Let E1, E2, E3 be the cylindrical frame field. For spherical coordinates, theunit vector field F2 in the direction of increasing J is the same as above, soF2 = E2. The unit vector field F1, in the direction of increasing r, pointsstraight out from the origin; hence it can be expressed as

(Fig. 2.21). Similarly, the vector field for increasing j is

F E E1 1 3= +cos sinj j

E U3 3= .

E U U2 1 2= - +sin cosJ J

86 2. Frame Fields

FIG. 2.19 FIG. 2.20

FIG. 2.21

Thus the formulas for E1, E2, E3 in (1) yield

By repeated use of the identity sin2 + cos2 = 1, we check that F1, F2, F3 isa frame field—the spherical frame field on R3. (Its actual domain of defini-tion is R3 minus the z axis, as in the cylindrical case.)

The following useful result is an immediate consequence of orthonormalexpansion.

6.3 Lemma Let E1, E2, E3 be a frame field on R3.(1) If V is a vector field on R3, then V = fiEi, where the functions

fi = V • Ei are called the coordinate functions of V with respect to E1,E2, E3.

(2) If V = fiEi and W = giEi, then V • W = figi. In particular,� V � = ( fi

2)1/2.

Thus a given vector field V has a different set of coordinate functions withrespect to each choice of a frame field E1, E2, E3. The Euclidean coordinatefunctions (Lemma 2.5 of Chapter 1), of course, come from the natural framefield U1, U2, U3. In Chapter 1, we used this natural frame field exclusively, butnow we shall gradually shift to arbitrary frame fields. The reason is clear: In studying curves and surfaces in R3, we shall then be able to choose a framefield specifically adapted to the problem at hand. Not only does this simplifycomputations, but it gives a clearer understanding of geometry than if wehad insisted on using the same frame field in every situation.

Exercises

1. If V and W are vector fields on R3 that are linearly independent at eachpoint, show that

is a frame field, where W = W - (W • E1)E1.

EVV

EW

WE E E1 2 3 1 2= = = ¥, ,

˜

˜

ÂÂÂÂ

Â

F U U U

F U U

F U U U

1 1 2 3

2 1 2

3 1 2 3

= +( ) +

= - +

= - +( ) +

cos cos sin sin

sin cos

sin cos sin cos .

j J J j

J J

j J J j

,

,

F E E3 1 3= - +sin cos .j j

2.6 Frame Fields 87

2. Express each of the following vector fields (i) in terms of the cylindricalframe field (with coefficients in terms of r, J, z) and (ii) in terms of the spher-ical frame field (with coefficients in terms of r, J, j):

(a) U1. (b) cosJU1 + sinJU2 + U3.(c) xU1 + yU2 + zU3.

3. Find a frame field E1, E2, E3 such that

2.7 Connection Forms

Once more we state the essential point: The power of the Frenet formulasstems not from the fact that they tell what the derivatives T ¢, N¢, B¢ are, butfrom the fact that they express these derivatives in terms of T, N, B—andthereby define curvature and torsion. We shall now do the same thing withan arbitrary frame field E1, E2, E3 on R3; namely, express the covariant deriv-atives of these vector fields in terms of the vector fields themselves. We beginwith the covariant derivative with respect to an arbitrary tangent vector v ata point p. Then

where by orthonormal expansion the coefficients of these equations are

These coefficients cij, depend on the particular tangent vector v, so a betternotation for them is

Thus for each choice of i and j, wij is a real-valued function defined on alltangent vectors. But we have met that kind of function before.

7.1 Lemma Let E1, E2, E3 be a frame field on R3. For each tangent vectorv to R3 at the point p, let

w ij v i jE E i jv p( ) = — ( ) ( )• ., ,1 3� �

w ij v i jE E i jv p( ) = — ( ) ( )• ., ,1 3� �

c E E i jij v i j= — ( )• .p for ,1 3� �

— = ( ) + ( ) + ( )— = ( ) + ( ) + ( )— = ( ) + ( ) + ( )

v

v

v

E c E c E c E

E c E c E c E

E c E c E c E

1 11 1 12 2 13 3

2 21 1 22 2 23 3

3 31 1 32 2 33 3

p p p

p p p

p p p

,

,

,

E x U x z U x z U1 1 2 3= + +cos sin cos sin sin .

88 2. Frame Fields

Then each wij is a 1-form, and wij = -wji. These 1-forms are called the con-nection forms of the frame field E1, E2, E3.

Proof. By definition, wij is a real-valued function on tangent vectors, soto verify that wij is a 1-form (Def. 5.1 of Ch. 1), it suffices to check the lin-earity condition. Using Theorem 5.3, we get

To prove that wij = -wji we must show that wij(v) = -wji(v) for everytangent vector v. By definition of frame field, Ei • Ej = dij, and since eachKronecker delta has constant value 0 or 1, the Leibnizian formula (4) ofTheorem 5.3 yields

By the symmetry of the dot product, the two vectors in this last term maybe reversed, so we have found that 0 = wij(v) + wji(v). ◆

The geometric significance of the connection forms is no mystery. The def-inition wij(v) = —vEi • Ej(p) shows that wij(v) is the initial rate at which Ei

rotates toward Ej as p moves in the v direction. Thus the 1-forms wij containthis information for all tangent vectors to R3.

The following basic result is little more than a rephrasing of the definitionof connection forms.

7.2 Theorem Let wij (1 � i, j � 3) be the connection forms of a framefield E1, E2, E3 on R3. Then for any vector field V on R3,

We call these the connection equations of the frame field E1, E2, E3.

Proof. For fixed i, both sides of this equation are vector fields. Thus wemust show that at each point p,

— = ( )( ) ( )( ) ÂV p i ij jj

E V Ew p p .

— = ( ) ( )ÂV i ij jj

E V E iw , 1 3� � .

0 = [ ] = — ( ) + ( ) —v p pE E E E E Ei j v i j i v j• • • .

w

w w

ij av bw i j

v i w i j

v i j w i j

ij ij

a b E E

a E b E E

a E E b E E

a b

v w p

p

p p

v w

+( ) = — ( )= — + —( ) ( )= — ( ) + — ( )= ( ) + ( )

+ •

•

• •

.

2.7 Connection Forms 89

But as we have already seen, the very definition of connection form makesthis equation a consequence of orthonormal expansion. ◆

When i = j, the skew-symmetry condition wij = -wji becomes wii = -wii;thus

Hence this condition has the effect of reducing the nine 1-forms wij for 1 � i, j � 3 to essentially only three, say w12, w13, w23. It is perhaps best toregard the connection forms wij as the entries of a skew-symmetric matrix of1-forms,

Thus in expanded form, the connection equations (Theorem 7.2) become

(*)

showing an obvious relation to the Frenet formulas

The absence from the Frenet formulas of terms corresponding to w13(V)E3

and -w13(V)E1 is a consequence of the special way the Frenet frame field isfitted to its curve. Having gotten T(~E1), we chose N(~E2) so that the deriv-ative T ¢ would be a scalar multiple of N alone and not involve B(~E3).

Another difference between the Frenet formulas and the equations abovestems from the fact that R3 has three dimensions, while a curve has but one.The coefficients—curvature k and torsion t—in the Frenet formulas measurethe rate of change of the frame field T, N, B only along its curve, that is, inthe direction of T alone. But the coefficients in the connection equations mustbe able to make this measurement for E1, E2, E3 with respect to arbitraryvector fields in R3. This is why the connection forms are 1-forms and not justfunctions.

These formal differences aside, a more fundamental distinction stands out.It is because a Frenet frame field is specially fitted to its curve that the Frenet

¢ =¢ = - +¢ = -

T N

N N B

B N

kk t

t

,

,

.

— = ( ) + ( )— = - ( ) + ( )— = - ( ) - ( )

V

V

V

E V E V E

E V E V E

E V E V E

1 12 2 13 3

2 12 1 23 3

3 13 1 23 2

,

,

.

w ww w

w w

ww w ww w ww w w

w ww ww w

=Ê

Ë

ÁÁ

ˆ

¯

˜˜

= -- -

Ê

Ë

ÁÁ

ˆ

¯

˜˜

11 12 13

21 22 23

31 32 33

12 13

12 23

13 23

0

0

0

.

w w w11 22 33 0= = = .

90 2. Frame Fields

formulas give information about that curve. Since the frame field E1, E2, E3

used above is completely arbitrary, the connection equations give no directinformation about R3, but only information about the “rate of rotation” ofthat particular frame field. This is not a weakness, but a strength, since asindicated earlier, if we can fit a frame field to a geometric problem arising inR3, then the connection equations will give direct information about thatproblem. Thus, these equations play a fundamental role in all the differentialgeometry of R3. For example, the Frenet formulas can be deduced from them(Exercise 8).

Given an arbitrary frame field E1, E2, E3 on R3, it is fairly easy to find anexplicit formula for its connection forms. First use orthonormal expansionto express the vector fields E1, E2, E3 in terms of the natural frame field U1,U2, U3 on R3:

Here each aij = Ei • Uj is a real-valued function on R3. The matrix

with these functions as entries is called the attitude matrix of the frame fieldE1, E2, E3. In fact, at each point p, the numerical matrix

is exactly the attitude matrix of the frame E1(p), E2(p), E3(p) as in Definition1.6. Since attitude matrices are orthogonal, the transpose tA of A is equal toits inverse A-1.

Define the differential of A = (aij) to be dA = (daij), so dA is a matrixwhose entries are 1-forms. We can now give a simple expression for the con-nection forms in terms of the attitude matrix.

7.3 Theorem If A = (aij) is the attitude matrix and w = (wij) the matrixof connection forms of a frame field E1, E2, E3, then

or equivalently,

w = ( )dA At matrix multiplication ,

A aijp p( ) = ( )( )

A a

a a a

a a a

a a aij= ( ) =

Ê

Ë

ÁÁ

ˆ

¯

˜˜

11 12 13

21 22 23

31 32 33

E a U a U a U

E a U a U a U

E a U a U a U

1 11 1 12 2 13 3

2 21 1 22 2 23 3

3 31 1 32 2 33 3

= + +

= + +

= + +

,

,

.


Since the proof is routine, it may be more informative to illustrate the resultby an example. For the cylindrical frame field in Example 6.2, we found theattitude matrix

Thus

Since w12 = dJ is the only nonzero connection form (except, of course,w21 = -w12), the connection equations (*) reduce to

These equations have immediate geometrical significance. Because V isarbitrary, the third equation says that the vector field E3 is parallel. We knewthis already since in the cylindrical frame field, E3 is just U3.

The first two equations tell us that the covariant derivatives of E1 and E2

with respect to a vector field V depend only on the rate of change of the angleJ in the V direction.

For example, the definition of J shows that V [J] = 0 whenever V is avector field that at each point is tangent to a plane through the z axis. Thusfor a vector field of this type the connection equations above predict that —VE1 = —VE2 = 0. In fact, it is clear from Fig. 2.19 that E1 and E2 do remainparallel on any plane through the z axis.

— = ( ) = [ ]— = - ( ) = - [ ]— =

V

V

V

E d V E V E

E d V E V E

E

1 2 2

2 1 1

3 0

J J

J J

,

,

.

wJ J J JJ J J J

J JJ J

JJ

= =-- -

Ê

Ë

ÁÁ

ˆ

¯

˜˜

-Ê

Ë

ÁÁ

ˆ

¯

˜˜

= -Ê

Ë

ÁÁ

ˆ

¯

˜˜

dA A

d d

d d

d

d

t

sin cos

cos sin

cos sin

sin cos

.

0

0

0 0 0

0

0

0 0 1

0 0

0 0

0 0 0

A = -Ê

Ë

ÁÁ

ˆ

¯

˜˜

cos sin

sin cos .

J JJ J

0

0

0 0 1

w ij jk ikk

a da i j= Â for ,1 3� � .

92 2. Frame Fields

Exercises

1. For any function f, show that the vector fields

form a frame field, and find its connection forms.

2. Find the connection forms of the natural frame field U1, U2, U3.

3. For any function f, show that

is the attitude matrix of a frame field, and compute its connection forms.

4. Prove that the connection forms of the spherical frame field are

5. If E1, E2, E3 is a frame field and W = fiEi, prove the covariant deriva-tive formula:

6. Let E1, E2, E3 be the cylindrical frame field. If V is a vector field suchthat V[r] = r and V[J] = 1, compute —V (r cosJE1 + r sinJE3).

7. (Computer.) (a) Write a computer command that, given the attitudematrix A of a frame field on R3, returns the matrix w = dA tA of its connection forms. (Hint: For Maple, use the differential operator d from the package difforms. For Mathematica, use the total differential Dt.) (b) Test part (a) on the cylindrical frame field and on the spherical frame field(Ex. 4).

8. Let b be a unit-speed curve in R3 with k > 0, and suppose that E1, E2, E3

is a frame field on R3 such that the restriction of these vector fields to b givesthe Frenet-frame field T, N, B of b. Prove that

w k w w t12 13 230T T T( ) = ( ) = ( ) =, , .

— = [ ] + ( )ÏÌÓ

¸˛ÂÂV i i ij

ij

j

W V f f V Ew .

Â

w j J w j w j J12 13 23= = =cos sin .d d d, ,

A

f f f f

f f f f

f f

= --

Ê

Ë

ÁÁ

ˆ

¯

˜˜

cos cos sin sin

sin cos sin cos

sin cos

2

2

0

E f U U f U

E f U U f U

E f U f U

1 1 2 3

2 1 2 3

3 1 3

2

2

= + -( )

= - -( )= +

sin cos

sin cos

cos sin

,

,


Then deduce the Frenet formulas from the connection equations. (Hint:Ex. 5 of Sec. 5.)

2.8 The Structural Equations

We have seen that 1-forms—the connection forms—give the simplest descrip-tion of the rate of rotation of a frame field. Furthermore, the frame fielditself can be described in terms of 1-forms.

8.1 Definition If E1, E2, E3 is a frame field on R3, then the dual 1-formsq1, q2, q3 of the frame field are the 1-forms such that

for each tangent vector v to R3 at p.

Note that qi is linear on the tangent vectors at each point; hence it is a 1-form. In particular, qi(Ej) = dij, so readers familiar with the notion of dualvector spaces will recognize that at each point, q1, q2, q3 gives the dual basisof E1, E2, E3.

In the case of the natural frame field U1, U2, U3, the dual forms are justdx1, dx2, dx3. In fact, from Example 5.3 of Chapter 1 we get

for each tangent vector v; hence dxi = qi.Using dual forms, the orthonormal expansion formula in Lemma 6.3 may

be written V = qi(V )Ei. In the characteristic fashion of duality, thisformula becomes the following lemma.

8.2 Lemma Let q1, q2, q3 be the dual 1-forms of a frame field E1, E2, E3.Then any 1-form f on R3 has a unique expression

Proof. Two 1-forms are the same if they have the same value on anyvector field V. But

◆

f q f q

f q f

E V E V

V E V

i i i i

i i

( )( )( ) = ( ) ( )

= ( )( ) = ( )Â Â

Â .

f f q= ( )Â Ei i .

Â

dx v Ui i iv v p( ) = = ( )•

q i iEv v p( ) = ( )•

94 2. Frame Fields

Thus f is expressed in terms of dual forms of E1, E2, E3 by evaluating iton E1, E2, E3. This useful fact is the generalization to arbitrary frame fieldsof Lemma 5.4 of Chapter 1.

We compared a frame field E1, E2, E3 to the natural frame field by meansof its attitude matrix A = (aij), for which

The dual formulation is just

with the same coefficients. In fact, by the preceding lemma,

But

These formulas for Ei and qi show plainly that q1, q2, q3 is merely the dualdescription of the frame field E1, E2, E3.

In calculus, when a new function appears on the scene, it is natural to askwhat its derivative is. Similarly with 1-forms—having associated with eachframe field its dual forms and connection forms, it is reasonable to ask whattheir exterior derivatives are. The answer is given by two neat sets of equa-tions discovered by Cartan.

8.3 Theorem (Cartan structural equations.) Let E1, E2, E3 be a framefield on R3 with dual forms q1, q2, q3 and connection forms wij (1 � i, j � 3).The exterior derivatives of these forms satisfy

(1) the first structural equations:

(2) the second structural equations:

Because qi is the dual of Ei, the first structural equations may be easily rec-ognized as the dual of the connection equations. Only later experience willshow that the second structural equations mean that R3 is flat—roughlyspeaking, in the same sense that the plane R2 is flat.

d i jij ik kjk

w w w= Ÿ ( )Â 1 3� �, .

d ii ij jj

q w q= Ÿ ( )Â 1 3� � ;

q di j i j ik k j ik kj ijU E U a U U a a( ) = = ( ) = =Â Â• • .

q qi i j jU dx= ( )Â .

q i ij ja dx= Â

E a U ii ij j= £ £( )Â 1 3 .

2.8 The Structural Equations 95

The most efficient proof of the structural equations requires some prelim-inary remarks. In the Cartan approach, the fundamental objects are not indi-vidual forms, but rather matrices whose entries are forms. We have alreadyseen that the simplest description of the connection forms wij of a frame fieldis as a single skew-symmetric matrix w with entries wij. Then, for example, wis expressed in terms of the attitude matrix A of the frame field by the matrixequation w = dA tA. (Here, as always, to apply d to a matrix, apply it to eachentry of the matrix.)

Similarly, the dual forms of a frame field can be described by a single n ¥ 1matrix q with entries qi. If x is the n ¥ 1 matrix whose entries are the naturalcoordinates xi of R3, then

so the formula qi = aij dxj above can be written as

For such matrices of forms, matrix multiplication is defined as usual, butof course when entries are multiplied it is by the wedge product.

The proof of Theorem 8.3 is now quite simple. Recall that since the atti-tude matrix A is orthogonal, tAA is the identity matrix I, which can be insertedin any matrix formula without effect.

Proof of the First Structural Equation. Since d 2 = 0, we evidentlyhave d(dx) = 0, so

Expressed in terms of entries, this is indeed the version in (1) of Theorem8.3.

Proof of the Second Structural Equation. For functions f and g.

Thus, using the transpose rule t(AB) = tB tA, we get

where the last step uses the skew-symmetry of w. Again, in terms of entries,this is the version in (2) of Theorem 8.3. ◆

d d dA A dA d A dA A A dAt t t t tw w w ww= ( ) = - ◊ ( ) = - ◊ ( ) = - = ,

d df g d g df dg df df dg( ) = ( ) = Ÿ = - Ÿ .

d d A d dA d dA A A dtq x x x wq= ( ) = ◊ = ◊ = .

q x= A d .

Â

qqqq

x=Ê

Ë

ÁÁ

ˆ

¯

˜˜

=Ê

Ë

ÁÁ

ˆ

¯

˜˜

1

2

3

1

2

3

and ,d

dx

dx

dx

96 2. Frame Fields

8.4 Example Structural equations for the spherical frame field (Example6.2). The dual forms and connection forms are

Let us check, say, the first structural equation

Using the skew-symmetry wij = -wji and the general properties of formsdeveloped in Chapter 1, we get

(the latter since dJ Ÿ dJ = 0). The sum of these terms is, correctly,

Second structural equations involve only one wedge product. For example,since w11 = w22 = 0,

In this case,

which is the same as

To derive the expressions given above for the dual 1-forms, first computedx1, dx2, dx3 by differentiating the well-known equations

Then substitute in the formula qi = aij dxj, where A = (aij) is the atti-tude matrix from Example 6.2. This result, somewhat disguised, is derived in elementary calculus by a familiar plausibility argument: If at each point thespherical coordinates r, J, j are altered by increments dr, dJ, dj, then thesides of the resulting infinitesimal box (Fig. 2.22) are dr, r cosj dJ, r dj.These are exactly the formulas for q1, q2, q3.

Â

x

x

x

1

2

3

=

=

=

r j J

r j J

r J

cos cos

cos sin

sin .

,

,

d d d d d d dw j J j J j j J12 = ( ) = ( ) Ÿ = - Ÿcos cos sin .

w w j j J j j J13 32Ÿ = Ÿ -( ) = - Ÿd d d dsin sin .

d k kw w w w w12 1 13 322= Ÿ = ŸÂ .

d d d d dq r j r j3 = ( ) = Ÿ .

w q j r r j

w q j J r j J

31 1

32 2 0

Ÿ = - Ÿ = Ÿ

Ÿ = -( ) Ÿ ( ) =

d d d d

d d

,

sin cos

d j jq w q w q w q3 3 31 1 32 2= Ÿ = Ÿ + ŸÂ .

q r w j Jq r j J w jq r j w j J

1 12

2 13

3 23

= == == =

d d

d d

d d

, ,

, ,

,

cos

cos

sin .

2.8 The Structural Equations 97

The structural equations provide a powerful method for dealing with geo-metrical problems in R3: Select a frame field well adapted to the problem athand; find its dual 1-forms and connection forms; apply the structural equa-tions; interpret the results. We will use this method later to study the geom-etry of surfaces in R3.

Exercises

1. For a 1-form f = fiqi, prove

(Compare Ex. 5 of Sec. 7.)

2. Check all the structural equations of the spherical frame field.

3. For the cylindrical frame field E1, E2, E3.(a) Starting from the basic cylindrical equations x = r cosJ, y = r sinJ,z = z, show that the dual 1-forms are

(b) Deduce that E1[r] = 1, E2[J] = 1/r, E3[z] = 1 and that the other six possibilities E1[J], . . . are all zero.(c) For a function f(r, J, z), show that

q q J q1 2 3= = =dr r d dz, , .

d df fj i iji

jj

f w q= +ÏÌÓ

¸˛

ŸÂÂ .

Â

98 2. Frame Fields

FIG. 2.22

4. Frame fields on R2. Given a frame field E1, E2 on R2 there is an anglefunction y such that

(a) Express the connection form and dual 1-forms in terms of y and thenatural coordinates x, y.(b) What are the structural equations in this case? Check that the resultsin part (a) satisfy these equations.

(Hint: Defining E3 = U3 gives a frame field on R3.)

2.9 Summary

We have accomplished the aims set at the beginning of this chapter. The ideaof a moving frame has been expressed rigorously as a frame field—either ona curve in R3 or on an open set of R3 itself. In the case of a curve, we usedonly the Frenet frame field T, N, B of the curve. Expressing the derivativesof these vector fields in terms of the vector fields themselves, we discoveredthe curvature and torsion of the curve. It is already clear that curvature andtorsion tell a lot about the geometry of a curve; we shall find in Chapter 3that they tell everything. In the case of an open set of R3, we dealt with anarbitrary frame field E1, E2, E3. Cartan’s generalization (Theorem 7.2) of theFrenet formulas followed the same pattern of expressing the (covariant)derivatives of these vector fields in terms of the vector fields themselves.Omitting the vector field V from the notation in Theorem 7.2, we have

Cartan’s equations are not conspicuously more complicated than Frenet’s,because the notion of 1-form is available for the coefficients wij, the connec-tion forms.

Cartan Frenet

E E E

E E E

E E E

T N

N T B

B N

— = +— = - +— = - -

¢ =¢ = - +¢ = -

1 12 2 13 3

2 12 1 23 3

3 13 1 23 2

w ww ww w

kk t

t

,

,

,

,

,

.

E U U

E U U

1 1 2

2 1 2

= +

= - +

cos sin ,

sin cos .

y y

y y

E ffr

E fr

fE f

fz1 2 3, ,[ ] =

∂∂

[ ] =∂∂

[ ] =∂∂

1J

.

2.9 Summary 99

▼

▲Chapter 3

Euclidean Geometry

100

We recall some familiar features of plane geometry. First of all, two trian-gles are congruent if there is a rigid motion of the plane that carries one tri-angle exactly onto the other. Corresponding angles of congruent triangles areequal, corresponding sides have the same length, the areas enclosed are equal,and so on. Indeed, any geometric property of a given triangle is automati-cally shared by every congruent triangle. Conversely, there are a number ofsimple ways in which one can decide whether two given triangles are con-gruent—for example, if for each the same three numbers occur as lengths ofsides.

In this chapter we shall investigate the rigid motions (isometries) of Euclid-ean space, and see how these remarks about triangles can be extended to othergeometric objects.

3.1 Isometries of R3

An isometry, or rigid motion, of Euclidean space is a mapping that preservesthe Euclidean distance d between points (Definition 1.2, Chapter 2).

1.1 Definition An isometry of R3 is a mapping F: R3 Æ R3 such that

for all points p, q in R3.

1.2 Example (1) Translations. Fix a point a in R3 and let T be the mapping that adds a to every point of R3. Thus T(p) = p + a for all

d F F dp q p q( ) ( )( ) = ( ), ,

3.1 Isometries of R3 101

points p. T is called translation by a. It is easy to see that T is an isometry,since

(2) Rotation around a coordinate axis. A rotation of the xy plane throughan angle J carries the point (p1, p2) to the point (q1, q2) with coordinates (Fig. 3.1)

Thus a rotation C of three-dimensional Euclidean space R3 around the z axis,through an angle J, has the formula

Evidently, the mapping C is a linear transformation. A straightforward com-putation shows that C preserves Euclidean distance, so it is an isometry.

Recall that if F and G are mappings of R3, the composite function GF isa mapping of R3 obtained by applying first F, then G.

1.3 Lemma If F and G are isometries of R3, then the composite mappingGF is also an isometry of R3.

Proof. Since G is an isometry, the distance from G(F(p)) to G(F(q)) isd(F(p), F(q)). But since F is an isometry, this distance equals d(p, q). ThusGF preserves distance; hence it is an isometry. ◆

C C p p p p p p p pp( ) = ( ) = - +( )1 2 3 1 2 1 2 3, , , ,cos sin sin cos .J J J J

q p p2 1 2= +sin cos .J J

q p p1 1 2= -cos sinJ J,

= - = ( )p q p qd , .

= +( ) - +( )p a q a

d T T dp q p a q a( ) ( )( ) = + +( ), ,

FIG. 3.1

102 3. Euclidean Geometry

In short, a composition of isometries is again an isometry.We also recall that if F: R3 Æ R3 is both one-to-one and onto, then F has

a unique inverse function F -1: R3 Æ R3, which sends each point F(p) back top. The relationship between F and F -1 is best described by the formulas

where I is the identity mapping of R3, that is, the mapping such that I(p) = pfor all p.

Translations of R3 (as defined in Example 1.2) are the simplest type ofisometry.

1.4 Lemma (1) If S and T are translations, then ST = TS is also a translation.

(2) If T is translation by a, then T has an inverse T -1, which is translationby -a.

(3) Given any two points p and q of R3, there exists a unique translationT such that T(p) = q.

Proof. To prove (3), for example, note that translation by q - p certainlycarries p to q. This is the only possibility, since if T is translation by a andT(p) = q, then p + a = q; hence a = q - p. ◆

A useful special case of (3) is that if T is a translation such that for someone point T(p) = p, then T = I.

The rotation in Example 1.2 is an example of an orthogonal transformationof R3, that is, a linear transformation C: R3 Æ R3 that preserves dot productsin the sense that

1.5 Lemma If C: R3 Æ R3 is an orthogonal transformation, then C is anisometry of R3.

Proof. First we show that C preserves norms. By definition, ||p||2 = p • p;hence

Thus || C(p) || = || p || for all points p. Since C is linear, it follows easily thatC is an isometry:

C C Cp p p p p p( ) = ( ) ( ) = =2 2• • .

C Cp q p q p q( ) ( ) =• • .for all ,

FF I F F I- -= =1 1, ,


◆

Our goal now is Theorem 1.7, which asserts that every isometry can beexpressed as an orthogonal transformation followed by a translation. Themain part of the proof is the following converse of Lemma 1.5.

1.6 Lemma If F is an isometry of R3 such that F(0) = 0, then F is anorthogonal transformation.

Proof. First we show that F preserves dot products; then we show thatF is a linear transformation. Note that by definition of Euclidean distance,the norm || p || of a point p is just the Euclidean distance d(0, p) from theorigin to p. By hypothesis, F preserves Euclidean distance, and F(0) = 0;hence

Thus F preserves norms. Now by a standard trick (“polarization”), we shalldeduce that it also preserves dot products. Since F is an isometry,

for any pair of points. Hence

By the definition of norm, this implies

Hence

The norm terms here cancel, since F preserves norms, and we find

as required.It remains to prove that F is linear. Let u1, u2, u3 be the unit points

(1, 0, 0), (0, 1, 0), (0, 0, 1), respectively. Then we have the identity

p u= ( ) = Âp p p pi i1 2 3, , .

F Fp q p q( ) ( ) =• • ,

F F F Fp p q q p p q q( ) - ( ) ( ) + ( ) = - +2 2 2 22 2• • .

F F F Fp q p q p q p q( ) - ( )( ) ( ) - ( )( ) = -( ) -( )• • .

F Fp q p q( ) - ( ) = - .

d F F dp q p q( ) ( )( ) = ( ), ,

F d F d F F dp 0 p 0 p 0 p p( ) = ( )( ) = ( ) ( )( ) = ( ) =, , , .

= ( )d p q p q, for all , .

d C C C C Cp q p q p q p q( ) ( )( ) = ( ) - ( ) = -( ) = -,


Also, the points u1, u2, u3 are orthonormal; that is, ui • uj = dij.We know that F preserves dot products, so F(u1), F(u2), F(u3) must also

be orthonormal. Thus orthonormal expansion gives

But

so

Using this identity, it is a simple matter to check the linearity condition

◆

We now give a concrete description of an arbitrary isometry.

1.7 Theorem If F is an isometry of R3, then there exist a unique trans-lation T and a unique orthogonal transformation C such that

Proof. Let T be translation by F(0). Then Lemma 1.4 shows that T -1 istranslation by -F(0). But T -1 F is an isometry, by Lemma 1.3, and furthermore,

Thus by Lemma 1.6, T -1 F is an orthogonal transformation, say T -1F = C.Applying T on the left, we get F = TC.

To prove the required uniqueness, we suppose that F can also be expressedas , where is a translation and an orthogonal transformation. Wemust prove = T and = C. Now TC = ; hence C = T -1 . Since Cand are linear transformations, they of course send the origin to itself. Itfollows that (T -1 )(0) = 0. But since T -1 is a translation, we conclude thatT -1 = I; hence = T. Then the equation TC = becomes TC = T .Applying T -1 gives C = .. ◆

Thus every isometry of R3 can be uniquely described as an orthogonal trans-formation followed by a translation. When F = TC as in Theorem 1.7, we call

CCCTTT

TTC

CTCTCTCTCT

T F T F F F- -( )( ) = ( )( ) = ( ) - ( ) =1 10 0 0 0 0.

F TC= .

F a b aF bFp q p q+( ) = ( ) + ( ).

F p Fi ip u( ) = ( )Â .

F F pi i ip u p u( ) ( ) = =• • ,

F F F Fi ip p u u( ) = ( ) ( ) ( )Â • .


C the orthogonal part of F, and T the translation part of F. Note that CT isgenerally not the same as TC (Exercise 1).

This decomposition theorem is the decisive fact about isometries of R3 (andits proof holds for Rn as well). We will use it to find an explicit formula foran arbitrary isometry.

First, recall from linear algebra that if C: R3 Æ R3 is any linear transfor-mation, its matrix (relative to the natural basis of R3) is the 3¥3 matrix {cij}such that

Thus, using the column-vector conventions, q = C(p) can be written as

By a standard result of linear algebra, a linear transformation of R3 isorthogonal (preserves dot products) if and only if its matrix is orthogonal(transpose equals inverse).

Returning to the decomposition F = TC in Theorem 1.7, if T is transla-tion by a = (a1, a2, a3), then

Using the above formula for C(p), we get

Alternatively, using the column-vector conventions, q = F(p) means

Exercises

Throughout these exercises, A, B, and C denote orthogonal transformations(or their matrices), and Ta is translation by a.

1. Prove that CTa = TC(a)C.

2. Given isometries F = TaA and G = TbB, find the translation and orthog-onal part of FG and GF.

q

q

q

a

a

a

c c c

c c c

c c c

p

p

p

1

2

3

1

2

3

11 12 13

21 22 23

31 32 33

1

2

3

Ê

Ë

ÁÁ

ˆ

¯

˜˜

=Ê

Ë

ÁÁ

ˆ

¯

˜˜

+Ê

Ë

ÁÁ

ˆ

¯

˜˜

Ê

Ë

ÁÁ

ˆ

¯

˜˜.

F F p p p a c p a c p a c pj j j j j jp( ) = ( ) = + + +( )Â Â Â1 2 3 1 1 2 2 3 3, , , , .

F TC Cp p a p( ) = ( ) = + ( ).

q

q

q

c c c

c c c

c c c

p

p

p

1

2

3

11 12 13

21 22 23

31 32 33

1

2

3

Ê

Ë

ÁÁ

ˆ

¯

˜˜

=Ê

Ë

ÁÁ

ˆ

¯

˜˜

Ê

Ë

ÁÁ

ˆ

¯

˜˜.

C p p p c p c p c pj j j j j j1 2 3 1 2 3, , ,( ) = ( )Â Â Â, .


3. Show that an isometry F = TaC has an inverse mapping F -1, which isalso an isometry. Find the translation and orthogonal parts of F -1.

4. If

show that C is orthogonal; then compute C(p) and C(q), and check that C(p) • C(q) = p • q.

5. Let F = TaC, where a = (1, 3, -1) and

If p = (2, -2, 8), find the coordinates of the point q for which(a) q = F(p). (b) q = F -1(p).(c) q = (CTa) (p).

6. In each case decide whether F is an isometry of R3. If so, find its trans-lation and orthogonal parts.

(a) F(p) = -p. (b) F(p) = (p • a) a, where || a || = 1.(c) F(p) = (p3 - 1, p2 - 2, p1 - 3). (d) F(p) = (p1, p2, 1).

A group G is a set furnished with an operation that assigns to each pair g1, g2

of elements of G an element g1g2, subject to these rules: (1) associative law:(g1g2)g3 = g1(g2g3), (2) there is a unique identity element e such that eg =ge = g for all g in G, and (3) inverses: For each g in G there is an element g-1 in G such that gg-1 = g-1 g = e.

Groups occur naturally in many parts of geometry, and we shall mentiona few in subsequent exercises. Basic properties of groups may be found in avariety of elementary textbooks.

7. Prove that the set E(3) of all isometries of R3 forms a group—with com-position of functions as the operation. E(3) is called the Euclidean group oforder 3.

A subset H of a group G is a subgroup of G provided (1) if g1 and g2 are inH, then so is g1g2, (2) is g is in H, so is g-1, and hence (3) the identity elemente of G is in H. A subgroup H of G is automatically a group.

C =-Ê

Ë

ÁÁÁ

ˆ

¯

˜˜

1 2 0 1 2

0 1 0

1 2 0 1 2

.

C =- -

-Ê

Ë

ÁÁ

ˆ

¯

˜˜

= -( )= ( )

ÏÌÓ

2 3 2 3 1 3

2 3 1 3 2 3

1 3 2 3 2 3

3 1 6

1 0 3and

, , ,

, , ,

p

q

3.2 The Tangent Map of an Isometry 107

8. Prove that the set T (3) of all translations of R3 and the set O(3) of allorthogonal transformations of R3 are each subgroups of the Euclidean groupE(3). O(3) is called the orthogonal group of order 3. Which isometries of R3

are in both these subgroups?

It is easy to check that the results of this section, though stated for R3, remainvalid for Euclidean spaces Rn of any dimension.

9. (a) Give an explicit description of an arbitrary 2 ¥ 2 orthogonal matrixC. (Hint: Use an angle and a sign.)

(b) Give a formula for an arbitrary isometry F of R = R1.

3.2 The Tangent Map of an Isometry

In Chapter 1 we showed that an arbitrary mapping F: R3 Æ R3 has a tangentmap F* that carries each tangent vector v at p to a tangent vector F*(v) atF(p). If F is an isometry, its tangent map is remarkably simple. (Since the dis-tinction between tangent vector and point is crucial here, we temporarilyrestore the point of application to the notation.)

2.1 Theorem Let F be an isometry of R3 with orthogonal part C.Then

for all tangent vectors vp to R3.Verbally: To get F*(vp), first shift the tangent vector vp to the canonically

corresponding point v of R3, then apply the orthogonal part C of F, andfinally shift this point C(v) to the canonically corresponding tangent vectorat F(p) (Fig. 3.2). Thus all tangent vectors at all points p of R3 are “rotated”in exactly the same way by F*—only the new point of application F(p) dependson p.

Proof. Write F = TC as in Theorem 1.7. Let T be translation by a, soF(p) = a + C(p). If vp is a tangent vector to R3, then by Definition 7.4 ofChapter 1, F*(vp) is the initial velocity of the curve t Æ F(p + tv). But usingthe linearity of C, we obtain

= ( ) + ( )F tCp v .

F t TC t T C tC C tCp v p v p v a p v+( ) = +( ) = ( ) + ( )( ) = + ( ) + ( )

F Cp F p* v v( ) = ( ) ( )


Thus F*(vp) is the initial velocity of the curve t Æ F(p) + tC(v), which isprecisely the tangent vector C(v)F(p). ◆

Expressed in terms of Euclidean coordinates, this result becomes

where C = (cij) is the orthogonal part of the isometry F, and if Ui is evalu-ated at p, then is evaluated at F(p).

2.2 Corollary Isometries preserve dot products of tangent vectors. Thatis, if vp and wp are tangent vectors to R3 at the same point, and F is an isom-etry, then

Proof. Let C be the orthogonal part of F, and recall that C, being anorthogonal transformation, preserves dot products in R3. By Theorem 2.1,

where we have twice used Definition 1.3 of Chapter 2 (dot products oftangent vectors). ◆

Since dot products are preserved, it follows automatically that derived con-cepts such as norm and orthogonality are preserved. Explicitly, if F is an isom-etry, then || F*(v) || = || v ||, and if v and w are orthogonal, so are F*(v) and F*(w).Thus frames are also preserved: if e1, e2, e3 is a frame at some point p of R3 andF is an isometry, then F*(e1), F*(e2), F*(e3) is a frame at F(p). (A direct proof iseasy: ei • ej = dij, so by Corollary 2.2, F*(ei) • F*(ej) = ei • ej = dij.)

= =v w v w• •p p

F F C C C Cp p F p F p* • * • •v w v w v w( ) ( ) = ( ) ( ) = ( ) ( )( ) ( )

F Fp p p p* • * • .v w v w( ) ( ) =

Ui

F v U c v Uj jj

ij j ii j

* Â ÂÊËÁ

ˆ¯ = ,

,

FIG. 3.2

3.2 The Tangent Map of an Isometry 109

Assertion (3) of Lemma 1.4 shows how two points uniquely determine atranslation. We now show that two frames uniquely determine an isometry.

2.3 Theorem Given any two frames on R3, say e1, e2, e3 at the point pand f1, f2, f3 at the point q, there exists a unique isometry F of R3 such thatF*(ei) = fi for 1 � i � 3.

Proof. First we show that there is such an isometry. Let e1, e2, e3, and f1,f2, f3 be the points of R3 canonically corresponding to the vectors in the two frames. Let C be the unique linear transformation of R3 such thatC(ei) = f i for 1 � i � 3. It is easy to check that C is orthogonal. Then letT be a translation by the point q - C(p). Now we assert that the isometryF = TC carries the e frame to the f frame. First note that

Then using Theorem 2.1 we get

for 1 � i � 3.To prove uniqueness, we observe that by Theorem 2.1 this choice of C

is the only possibility for the orthogonal part of the required isometry. Thetranslation part is then completely determined also, since it must carry C(p)to q. Thus the isometry F = TC is uniquely determined. ◆

To compute the isometry in the theorem, recall that the attitude matrix Aof the e frame has the Euclidean coordinates of ei as its ith row: ai1, ai2, ai3.The attitude matrix B of the f frame is similar. We claim that C in the theorem(or strictly speaking, its matrix) is tBA. To verify this it suffices to check thattBA(ei) = fi, since this uniquely characterizes C. For i = 1 we find, using thecolumn-vector conventions,

that is, tBA(e1) = f1. The cases i = 2, 3 are similar; hence C = tBA. As notedabove, T is then necessarily translated by q - C(p).

=Ê

Ë

ÁÁ

ˆ

¯

˜˜

=Ê

Ë

ÁÁ

ˆ

¯

˜˜

Ê

Ë

ÁÁ

ˆ

¯

˜˜

=Ê

Ë

ÁÁ

ˆ

¯

˜˜

tB

b b b

b b b

b b b

b

b

b

1

0

0

1

0

0

11 21 31

12 22 32

13 23 33

11

12

13

,

t tBA

a

a

a

B

a a a

a a a

a a a

a

a

a

11

12

13

11 12 13

21 22 23

31 32 33

11

12

13

Ê

Ë

ÁÁ

ˆ

¯

˜˜

=Ê

Ë

ÁÁ

ˆ

¯

˜˜

Ê

Ë

ÁÁ

ˆ

¯

˜˜

F Ci i F p i F p i q i* ˆ ˆ ê e f f f( ) = ( ) = ( ) = ( ) =( ) ( )

F T C C Cp p q p p q( ) = ( )( ) = - ( ) + ( ) = .


Exercises

1. If T is a translation, show that for every tangent vector v the vector T(v)is parallel to v (same Euclidean coordinates).

2. Prove the general formulas (GF )* = G*F* and (F -1)* = (F*)-1 in thespecial case where F and G are isometries of R3.

3. Given the frame

at p = (0, 1, 0) and the frame

at q = (3, -1, 1), find a and C such that the isometry F = TaC carries the eframe to the f frame.

4. (a) Prove that an isometry F = TC carries the plane through p orthog-onal to q π 0 to the plane through F(p) orthogonal to C(q).

(b) If P is the plane through (1/2, -1, 0) orthogonal to (0, 1, 0) find anisometry F = TC such that F(P) is the plane through (1, -2, 1) orthogonalto (1, 0, -1).

5. (Computer.)(a) Verify that both sets of vectors in Exercise 3 form frames by showingthat A tA = I for their attitude matrices.(b) Find the matrix C that carries each ei to fi, and check this for i = 1, 2,3.

3.3 Orientation

We now come to one of the most interesting and elusive ideas in geometry.Intuitively, it is orientation that distinguishes between a right-handed gloveand a left-handed glove in ordinary space. To handle this concept mathe-matically, we replace gloves by frames and separate all the frames on R3 intotwo classes as follows. Recall that associated with each frame e1, e2, e3 at apoint of R3 is its attitude matrix A. According to the exercises for Section 1of Chapter 2,

e e e1 2 3 1• det .¥ = = ±A

f f f1 2 31 0 1 0 1 0 1 0 1 2= ( ) = ( ) = -( ), , 2 , , , , , ,

e e e1 2 32 2 1 3 2 1 2 3 1 2 2 3= ( ) = -( ) = -( ), , , , , , , ,

3.3 Orientation 111

When this number is +1, we shall say that the frame e1, e2, e3 is positively ori-ented (or right-handed); when it is -1, the frame is negatively oriented (or left-handed).

We omit the easy proof of the following facts.

3.1 Remark (1) At each point of R3 the frame assigned by the naturalframe field U1, U2, U3 is positively oriented.

(2) A frame e1, e2, e3 is positively oriented if and only if e1 ¥ e2 = e3. Thusthe orientation of a frame can be determined, for practical purposes, by the“right-hand rule” given at the end of Section 1 of Chapter 2. Pictorially, theframe (P) in Fig. 3.3 is positively oriented, whereas the frame (N ) is nega-tively oriented. In particular, Frenet frames are always positively oriented,since by definition, B = T ¥ N.

(3) For a positively oriented frame e1, e2, e3, the cross products are

For a negatively oriented frame, reverse the vectors in each cross product.(One need not memorize these formulas—the right-hand rule will give themall correctly.)

Having attached a sign to each frame on R3, we next attach a sign to eachisometry F of R3. In Chapter 2 we proved the well-known fact that the deter-minant of an orthogonal matrix is either +1 or -1. Thus if C is the orthog-onal part of the isometry F, we define the sign of F to be the determinant ofC, with notation

sgn det .F C=

e e e e e3 1 2 2 1= ¥ = - ¥ .

e e e e e2 3 1 1 3= ¥ = - ¥ ,

e e e e e1 2 3 3 2= ¥ = - ¥ ,

FIG. 3.3


We know that the tangent map of an isometry carries frames to frames.The following result tells what happens to their orientations.

3.2 Lemma If e1, e2, e3 is a frame at some point of R3 and F is an isometry, then

Proof. If , then by the coordinate form of Theorem 2.1 wehave

where C = (cij) is the orthogonal part of F. Thus the attitude matrix of theframe F*(e1), F*(e2), F*(e3) is the matrix

But the triple scalar product of a frame is the determinant of its attitudematrix, and by definition, sgnF = detC. Consequently,

◆

This lemma shows that if sgnF = +1, then F* carries positively orientedframes to positively oriented frames and carries negatively oriented framesto negatively oriented frames. On the other hand, if sgnF = -1, positive goesto negative and negative to positive.

3.3 Definition An isometry F of R3 is said to be

where C is the orthogonal part of F.

orientation F C- ifreversing sgn det ,= = -1

orientation preserving F C- if sgn det ,= = +1

= ( ) ◊ ¥sgn .F e e e1 2 3

= ◊ = ◊det det det detC A C At

F F F C At* • * * dete e e1 2 3( ) ( ) ¥ ( ) = ( )

c a c a C Aik jkk

ikt

kjk

tÂ ÂÊË

ˆ¯ = Ê

Ëˆ¯ = .

F c a Uj ik jk ii k

* e( ) = Â ,,

e j jk ka U= Â

F F F F* • * * sgn • .e e e e e e1 2 3 1 2 3( ) ( ) ¥ ( ) = ( ) ¥

3.3 Orientation 113

3.4 Example (1) Translations. All translations are orientation-preserv-ing. Geometrically this is clear, and in fact the orthogonal part of a transla-tion T is just the identity mapping I, so sgnT = detI = +1.

(2) Rotations. Consider the orthogonal transformation C given in Example1.2, which rotates R3 through angle q around the z axis. Its matrix is

Hence sgnC = detC = +1, so C is orientation-preserving (see Exercise 4).

(3) Reflections. One can (literally) see reversal of orientation by using amirror. Suppose the yz plane of R3 is the mirror. If one looks toward thatplane, the point p = (p1, p2, p3) appears to be located at the point

(Fig. 3.4). The mapping R so defined is called reflection in the yz plane.Evidently it is an orthogonal transformation, with matrix

Thus R is an orientation-reversing isometry, as confirmed by the experimen-tal fact that the mirror image of a right hand is a left hand.

Both dot and cross product were originally defined in terms of Euclideancoordinates. We have seen that the dot product is given by the same formula,

v w e e• •= ( ) ( ) =Â Â Âv w v wi i i i i i ,

-Ê

Ë

ÁÁ

ˆ

¯

˜˜

1 0 0

0 1 0

0 0 1

.

R p p pp( ) = -( )1 2 3, ,

cos sin

sin cos .

q qq q

-Ê

Ë

ÁÁ

ˆ

¯

˜˜

0

0

0 0 1

FIG. 3.4


no matter what frame e1, e2, e3 is used to get coordinates for v and w. Almostthe same result holds for cross products, but orientation is now involved.

3.5 Lemma Let e1, e2, e3 be a frame at a point of R3. If and , then

where e = e1 • e2 ¥ e3 = ±1.

Proof. It suffices merely to expand the cross product

using the formulas (3) of Remark 3.1. For example, if the frame is posi-tively oriented, for the e1 component of v ¥ w we get

Since e = 1 in this case, we get the same result by expanding the determi-nant in the statement of this lemma. ◆

It follows immediately that the effect of an isometry on cross products alsoinvolves orientation.

3.6 Theorem Let v and w be tangent vectors to R3 at p. If F is an isometry of R3, then

Proof. Write . Now let

Since F* is linear,

A straightforward computation using Lemma 3.5 shows that

F F F* * *v w v w( ) ¥ ( ) = ¥( )e ,

F v F wi i i i* * .v e w e( ) = ( ) =Â Âand

e pi iF U= ( )( )* .

v p w p= ( ) = ( )Â ÂvU and wUi i i i

F F F F* sgn * * .v w v w¥( ) = ( ) ( ) ¥ ( )

v w v w v w v w2 2 3 3 3 3 2 2 2 3 3 2 1e e e e e¥ + ¥ = -( ) .

v w e e e e e e¥ = + +( ) ¥ + +( )v v v w w w1 1 2 2 3 3 1 1 2 2 3 3

v w

e e e

¥ = e1 2 3

1 2 3

1 2 3

v v v

w w w

,

w e= Âwi i

v e= Â vi i

3.3 Orientation 115

where

But U1, U2, U3 is positively oriented, so by Lemma 3.2, e = sgnF. ◆

Exercises

1. Prove

Deduce that sgn F = sgn (F -1).

2. If H0 is an orientation-reversing isometry of R3, show that everyorientation-reversing isometry has a unique expression H0F, where F is orientation-preserving.

3. Let v = (3, 1, -1) and w = (-3, -3, 1) be tangent vectors at some point.If C is the orthogonal transformation given in Exercise 4 of Section 1, checkthe formula

4. A rotation is an orthogonal transformation C such that det C = +1. Provethat C does, in fact, rotate R3 around an axis. Explicitly, given a rotation C,show that there exists a number J and points e1, e2, e3 with ei • ej = dij suchthat (Fig. 3.5)

C e e3 3( ) = .

C e e e2 1 2( ) = - +sin cosJ J ,

C e e e1 1 2( ) = +cos sinJ J ,

C C C C* sgn * * .v w v w¥( ) = ( ) ( ) ¥ ( )

sgn sgn sgn sgn .FG F G GF( ) = ◊ = ( )

e = ¥ = ( )( ) ( )( ) ¥ ( )( )e e e p p p1 2 3 1 2 3• * • * * .F U F U F U

FIG. 3.5


(Hint: The fact that the dimension of R3 is odd means that C has an eigen-value +1, so there is a point p π 0 such that C(p) = p.)

5. Let a be a point of R3 such that || a || = 1. Prove that the formula

defines an orthogonal transformation. Describe its general effect on R3.

6. Prove(a) The set O+(3) of all rotations of R3 is a subgroup of the orthogonalgroup O(3) (see Ex. 8 of Sec. 3.1).(b) The set E +(3) of all orientation-preserving isometries of R3 is a sub-group of the Euclidean group E(3).

3.4 Euclidean Geometry

In the discussion at the beginning of this chapter, we recalled a fundamentalfeature of plane geometry: If there is an isometry carrying one triangle ontoanother, then the two (congruent) triangles have exactly the same geometricproperties. A close examination of this statement will show that it does notadmit a proof—it is, in fact, just the definition of “geometric property of atriangle.” More generally, Euclidean geometry can be defined as the totalityof concepts that are preserved by isometries of Euclidean space. For example,Corollary 2.2 shows that the notion of dot product on tangent vectorsbelongs to Euclidean geometry. Similarly, Theorem 3.6 shows that the crossproduct is preserved by isometries (except possibly for sign).

This famous definition of Euclidean geometry is somewhat generous,however. In practice, the label “Euclidean geometry” is usually attached onlyto those concepts that are preserved by isometries, but not by arbitrary map-pings, or even the more restrictive class of mappings (diffeomorphisms) thatpossess inverse mappings. An example should make this distinction clearer.If a = (a1, a2, a3) is a curve in R3, then the various derivatives

look pretty much alike. Now, Theorem 7.8 of Chapter 1 asserts that velocityis preserved by arbitrary mappings F: R3 Æ R3, that is, if b = F(a), then b¢ =F*(a ¢). But it is easy to see that acceleration is not preserved by arbitrary map-pings. For example, if a(t) = (t, 0, 0) and F = (x2, y, z), then a≤ = 0; henceF*(a≤) = 0. But b = F(a) has the formula b(t) = (t2, 0, 0), so b≤ = 2U1. Thus

¢ = ÊË

ˆ¯ ¢¢ = Ê

Ëˆ¯a

a a aa

a a addt

ddt

ddt

ddt

ddt

ddt

1 2 32

1

2

22

2

23

2, , , ,, , . . .

C p a p p a a( ) = ¥ + ( )•

3.4 Euclidean Geometry 117

in this case, b = F(a), but b≤ π F*(a≤). We shall see in a moment, however,that acceleration is preserved by isometries.

For this reason, the notion of velocity belongs to the calculus ofEuclidean space, while the notion of acceleration belongs to Euclidean geom-etry. In this section we examine some of the concepts introduced in Chapter2 and prove that they are, in fact, preserved by isometries. (We leave largelyto the reader the easier task of showing that they are not preserved by diffeomorphisms.)

Recall the notion of vector field on a curve (Definition 2.2 of Chapter 2).If Y is a vector field on a: I Æ R3 and F: R3 Æ R3 is any mapping, then =F*(Y ) is a vector field on the image curve = F(a). In fact, for each t in I,Y(t) is a tangent vector to R3 at the point a(t). But then (t) = F*(Y(t)) is atangent vector to R3 at the point F(a(t)) = (t).

(These relationships are illustrated in Fig. 3.6.) Isometries preserve thederivatives of such vector fields.

4.1 Corollary Let Y be a vector field on a curve a in R3, and let F be anisometry of R3. Then = F*(Y ) is a vector field on = F(a), and

Proof. To differentiate a vector field , one simply differenti-ates its Euclidean coordinate functions, so

¢ = ÂYdydt

Ujj .

Y y Uj j= Â

¢ = ¢( )Y F Y* .

aY

aY

aY

FIG. 3.6


Thus by the coordinate version of Theorem 2.1, we get

On the other hand,

But each cij is constant, being by definition an entry in the matrix of theorthogonal part of the isometry F. Hence

Thus the vector fields F*(Y¢) and ¢ are the same. ◆

We claimed earlier that isometries preserve acceleration: If = F(a), whereF is an isometry, then ≤ = F*(a≤). This is an immediate consequence of thepreceding result, for if we set Y = a ¢, then by Theorem 7.8 of Chapter 1,

= ¢; hence

Now we show that the Frenet apparatus of a curve is preserved by isome-tries. This is certainly to be expected on intuitive grounds, since a rigid motionought to carry one curve into another that turns and twists in exactly the same way. And this is what happens when the isometry is orientation-preserving.

4.2 Theorem Let b be a unit-speed curve in R3 with positive cur-vature, and let = F(b) be the image curve of b under an isometry F of R3.Then

where sgnF = ±1 is the sign of the isometry F.

Proof. Note that is also a unit-speed curve, since

¢ = ¢( ) = ¢ =b b bF* .1

b

B F F B= ( ) ( )sgn * ,

t t= ( ) = ( )sgn *F N F N, ,

k k= = ( ), T F T* ,

b

¢¢ = ¢ = ¢( ) = ¢¢( )a aY F Y F* * .

aY

aa

Y

¢ = ( ) =Â ÂYddt

c y U cdydt

Uij j i ijj

i .

Y F Y c y Uij j i= ( ) = Â* .

F Y cdydt

Uijj

i* .¢( ) = Â

3.4 Euclidean Geometry 119

Thus the definitions in Section 3 of Chapter 2 apply to both b and , so

Since F* preserves both acceleration and norms, it follows from the definition of curvature that

To get the full Frenet frame, we now use the hypothesis k > 0 (whichimplies > 0, since = k). By definition, N = b≤/k ; hence using preced-ing facts, we find

It remains only to prove the interesting cases B and t. Since the defini-tion B = T ¥ N involves a cross product, we use Theorem 3.6 to get

The definition of torsion is essentially t = -B¢ • N = B • N¢. Thus, usingthe results above for B and N, we get

◆

The presence of sgn F in the formula for the torsion of F(b) shows thatthe torsion of a curve gives a more subtle description of the curve than hasbeen apparent so far. The sign of t measures the orientation of the twisting ofthe curve. If F is orientation-reversing, the formula = -t proves that thetwisting of the image of curve F(b) is exactly opposite to that of b itself.

A simple example will illustrate this reversal.

4.3 Example Let b be the unit-speed helix

gotten from Example 3.3 of Chapter 2 by setting a = b = 1; hence c = .We know from the general formulas for helices that k = t = 1/2. Now let Rbe reflection in the xy plane, so R is the isometry R(x, y, z) = (x, y, -z). Thusthe image curve = R(b) is the mirror image

b ssc

sc

sc

( ) = -ÊË

ˆ¯cos sin, ,

b

2

b ssc

sc

sc

( ) = ÊË

ˆ¯cos sin, , ,

t

t t= ¢ = ( ) ( ) ¢( ) = ( ) ¢ = ( )B N F F B F N F B N F• sgn * • * sgn • sgn .

B T N F T F N F F T N F F B= ¥ = ( ) ¥ ( ) = ( ) ¥( ) = ( ) ( )* * sgn * sgn * .

NF

F F N=¢¢

=¢¢( )

=¢¢Ê

Ëˆ¯ = ( )b

kb

kbk

** * .

kk

k b b b k= ¢¢ = ¢¢( ) = ¢¢ =F* .

T F F T= ¢ = ¢( ) = ( )b b* * .

b


of the original curve. One can see in Fig. 3.7 that the mirror has its usualeffect: b and twist in opposite ways—if b is “right-handed,” then is“left-handed.” (The fact that b is going up and down is, in itself, irrele-vant.) Formally: The reflection R is orientation-reversing; hence the theorempredicts and Since is just the helix gotten inExample 3.3 of Chapter 2 by taking a = 1 and b = -1, this may be checkedby the general formulas there.

Exercises

1. Let F = TC be an isometry of R3, b a unit speed curve in R3. Prove(a) If b is a cylindrical helix, then F(b) is a cylindrical helix.(b) If b has spherical image s, then F(b) has spherical image C(s).

2. Let Y = (t, 1 - t2, 1 + t2) be a vector field on the helix

and let C be the orthogonal transformation

Compute = C(a) and = C*(Y ), and check that

C Y Y C Y Y* * • • .¢( ) = ¢ ¢¢( ) = ¢¢ ¢ ¢¢ = ¢ ¢¢, ,a a a a

Ya

C =-

-Ê

Ë

ÁÁ

ˆ

¯

˜˜

1 0 0

0 1 2 1 2

0 1 2 1 2

.

a t t t t( ) = ( )cos sin, , ,2

bt t= - = - 12 .k k= = 1

2

bbb

FIG. 3.7

3.5 Congruence of Curves 121

3. Sketch the triangles in R2 that have vertices

Show that these triangles are congruent by exhibiting an isometry F =TC that carries D1 to D2. (Hint: the orthogonal part C is not altered if thetriangles are translated.)

4. If F: R3 Æ R3 is a diffeomorphism such that F* preserves dot products,show that F is an isometry. (Hint: Show that F preserves lengths of curve segments and deduce that F-1 does also.)

5. Let F be an isometry of R3. For each vector field V let be the vectorfield such that F*(V(p)) = (F(p)) for all p. Prove that isometries preservecovariant derivatives; that is, show =

3.5 Congruence of Curves

In the case of curves in R3, the general notion of congruence takes the fol-lowing form.

5.1 Definition Two curves a, b: I Æ E3 are congruent provided thereexists an isometry F of R3 such that b = F(a); that is, b(t) = F(a(t)) for all tin I.

Intuitively speaking, congruent curves are the same except for position inspace. They represent trips at the same speed along routes of the same shape.For example, the helix a(t) = (cos t, sin t, t) spirals around the z axis in exactlythe same way the helix b(t) = (t, cos t, sin t) spirals around the x axis. Evi-dently these two curves are congruent, since if F is the isometry such that

then F (a) = b.To decide whether given curves a and b are congruent, it is hardly practi-

cal to try all the isometries of R3 to see whether there is one that carries a tob. What we want is a description of the shape of a unit-speed curve so accu-rate that if a and b have the same description, then they must be congruent.The proper description, as the reader will doubtless suspect, is given by cur-vature and torsion. To prove this we need one preliminary result.

Curves whose congruence is established by a translation are said to be parallel. Thus, curves a, b: I Æ E3 are parallel if and only if there is a point

F p p p p p p1 2 3 3 1 2, , , , ,( ) = ( )

—V W.—V WV

V

D D1 23 1 7 1 7 4 2 0 2 5 2 5 16 5: : ., , , , , , , , , , ,( ) ( ) ( ) ( ) ( ) -( )


p in R3 such that b(s) = a(s) + p for all s in I, or, in functional notation,b = a + p.

5.2 Lemma Two curves a, b: I Æ R3 are parallel if their velocity vectorsa ¢(s) and b¢(s) are parallel for each s in I. In this case, if a(s0) = b(s0) for someone s0 in I, then a = b.

Proof. By definition, if a ¢(s) and b¢(s) are parallel, they have the sameEuclidean coordinates. Thus

where ai and bi are the Euclidean coordinate functions of a and b. But by elementary calculus, the equation dai/ds = dbi/ds implies that there is a constant pi such that bi = ai + pi. Hence b = a + p. Furthermore, ifa(s0) = b(s0), we deduce that p = 0; hence a = b. ◆

5.3 Theorem If a, b: I Æ R3 are unit-speed curves such that ka = kb andta = ±tb, then a and b are congruent.

Proof. There are two main steps:(1) Replace a by a suitably chosen congruent curve F(a).(2) Show that F(a) = b (Fig. 3.8).Our guide for the choice in (1) is Theorem 4.2. Fix a number, say 0, in

the interval I. If ta = tb, then let F be the (orientation-preserving) isome-try that carries the Frenet frame Ta(0), Na(0), Ba(0) of a at a(0) to the

dds

sdds

s ii ia b( ) = ( ) for ,1 3� �

FIG. 3.8


Frenet frame Tb(0), Nb(0), Bb(0), of b at b(0). (The existence of this isom-etry is guaranteed by Theorem 2.3.) Denote the Frenet apparatus of =F(a) by , , , , ; then it follows immediately from Theorem 4.2 andthe information above that

(‡)

On the other hand, if ta = -tb, we choose F to be the (orientation-reversing) isometry that carries Ta(0), Na(0), Ba(0) at a(0) to the frameTb(0), Nb(0), Bb(0) at b(0). (Frenet frames are positively oriented; hencethis last frame is negatively oriented: This is why F is orientation-reversing.) Then it follows from Theorem 4.2 that the equations (‡) holdalso for = F(a) and b. For example,

For step (2) of the proof, we shall show = Tb; that is, the unit tan-gents of = F(a) and b are parallel at each point. Since (0) = b(0), itwill follow from Lemma 5.2 that F(a) = b. On the interval I, consider thereal-valued function f = • Tb + • Nb + • Bb. Since these are unitvector fields, the Schwarz inequality (Sec. 1, Ch. 2) shows that

furthermore, • Tb = 1 if and only if = Tb. Similar remarks hold for theother two terms in f. Thus it suffices to show that f has constant value 3. By(‡), f(0) = 3. Now consider

A simple computation completes the proof. Substitute the Frenet for-mulas in this expression and use the equations = kb, = tb from (‡). Theresulting eight terms cancel in pairs, so f ¢ = 0, and f has, indeed, constantvalue 3. ◆

Thus, a unit-speed curve is determined but for position in R3 by its curvatureand torsion.

Actually the proof of Theorem 5.3 does more than establish that a and bare congruent; it shows how to compute explicitly an isometry carrying a tob. We illustrate this in a special case.

tk

¢ = ¢ + ¢ + ¢ + ¢ + ¢ + ¢f T T T T N N N N B B B B• • • • • •b b b b b b

TT

T T• ;b � 1

BNT

aaT

B F B B0 0 0( ) = - ( )( ) = ( )* .a b

a

t t b b= ( ) = ( ), .B B0 0

k k b b= ( ) = ( ), ,N N0 0

a b b0 0 0 0( ) = ( ) ( ) = ( ), ,T T

BNTtka


5.4 Example Consider the unit-speed curves a, b: R Æ R3 such that

where c = . Obviously, these curves are congruent by means of a reflec-tion—they are the helices considered in Example 4.3—but we shall ignorethis in order to describe a general method for computing the required isom-etry. According to Example 3.3 of Chapter 2, a and b have the same curva-ture, ka = 1/2 = kb ; but torsions of opposite sign, ta = 1/2 = -tb. Thus thetheorem predicts congruence by means of an orientation-reversing isometryF. From its proof we see that F must carry the Frenet frame

where a = 1/ , to the frame

where the minus sign will produce orientation reversal. (These explicit for-mulas also come from Example 3.3 of Chapter 2.) By the remark followingTheorem 2.3, the isometry F has orthogonal part C = tBA, where A and Bare the attitude matrices of the two frames above. Thus

since a = 1/ . These two frames have the same point of application a(0) =b(0) = (1, 0, 0). But C does not move this point, so the translation part of Fis just the identity map. Thus we have (correctly) found that the reflection F = C carries a to b.

From the viewpoint of Euclidean geometry, two curves in R3 are “thesame” if they differ only by an isometry of R3. What, for example, is a helix?

2

C a a

a a

a a

a a

=-

-- -

Ê

Ë

ÁÁ

ˆ

¯

˜˜

--

Ê

Ë

ÁÁ

ˆ

¯

˜˜

=-

Ê

Ë

ÁÁ

ˆ

¯

˜˜

0 1 0

0

0

0

1 0 0

0

1 0 0

0 1 0

0 0 1

- ( ) = - -( )B a ab 0 0, , ,

Nb 0 1 0 0( ) = -( ), , ,

T a ab 0 0( ) = -( ), , ,

2

B a aa 0 0( ) = -( ), , ,

Na 0 1 0 0( ) = -( ), , ,

T a aa 0 0( ) = ( ), , ,

2

b ssc

sc

sc

( ) = -ÊË

ˆ¯cos sin, , ,

a ssc

sc

sc

( ) = ÊË

ˆ¯cos sin, , ,


It is not just a curve that spirals around the z axis as in Example 3.3 ofChapter 2, but any curve congruent to one of these special helices. One cangive general formulas, but the best characterization follows.

5.5 Corollary Let a be a unit speed curve in R3. Then a is a helix if andonly if both its curvature and torsion are nonzero constants.

Proof. For any numbers a > 0 and b π 0, let ba,b be the special helix givenin Example 3.3 of Chapter 2. If a is congruent to ba,b, then (changing the sign of b if necessary) we can assume the isometry is orientation-preserving. Thus, a has curvature and torsion

Conversely, suppose a has constant nonzero k and t. Solving the pre-ceding equations, we get

Thus a and ba,b have the same curvature and torsion; hence they are congruent. ◆

Our results so far demand unit speed, but it is easy to weaken this restriction.

5.6 Corollary Let a, b: I Æ R3 arbitrary-speed curves. If

then the curves a and b are congruent.

The proof is immediate, for the data ensures that the unit speed parame-trizations of a and b have the same curvature and torsion—hence they arecongruent. But then the original curves are congruent under the same isometry since their speeds are the same.

The theory of curves we have presented applies only to regular curves withpositive curvature k > 0, because only for such curves is it possible to definethe Frenet frame field. However, an arbitrary curve a in R3 can be studied bymeans of an arbitrary frame field on a, that is, three unit-vector fields E1, E2,E3 on a that are orthogonal at each point.

v va b a b a bk k t t= > = > = ±0 0, , and ,

a b=+

=+

kk t

tk t2 2 2 2

, .

k t=+

=+

aa b

ba b2 2 2 2

, .


At a critical point later on, we will need this generalization of the congru-ence theorem (5.3):

5.7 Theorem Let a, b: I Æ R3 be curves defined on the same interval.Let E1, E2, E3 be a frame field on a, and F1, F2, F3 a frame field on b. If

(1) a ¢ • Ei = b¢ • Fi (1 � i � 3),(2) Ei¢ • Ej = Fi¢ • Fj (1 � i, j � 3),

then a and b are congruent.Explicitly, for any t0 in I, if F is the unique Euclidean isometry that sends

each Ei(t0) to Fi(t0), then F(a) = b.

Proof. Let F be the specified isometry. Since F* preserves dot products,it follows that the vector fields = F*(Ei) for 1 £ i £ 3 form a frame fieldon = F(a). And since F* preserves velocities of curves and derivativesof vector fields, by using condition (1) in the theorem, we find

(*)

Similarly, from condition (2), we get

(**)

In view of this last equation, orthonormal expansion yields

with the same coefficient functions aij. Note that aij + aji = 0; hence aii = 0.(Proof: Differentiate • = dij.)

Now let f = • Fi. We prove f = 3 as before: f(t0) = 3, and

Thus each • Fi = 1, that is, and Fi are parallel at each point. By (*)the same is true for

Since a(t0) = b(t0), Lemma 5.2 gives the required result, F(a) = = b.◆

5.8 Remark Existence theorem for curves in R3. Curvature and torsiontell whether two unit-speed curves are isometric, but they do more than that:

a

¢ = ¢( ) ¢ = ¢( )Â Âa a b b• • .E E F Fi i i iand

EiEi

¢ = ¢ + ¢( ) = +( ) =Â Âf E F E F a a E Fi i i ij ji j ii j

i• • • .0,

E jÂE jEi

¢ = ¢ =Â ÂE a E F a Fi ij jj

i ij jj

and ,

E t F t E E F F i ji i i j i j0 0 1 3( ) = ( ) ¢ = ¢ ◊and for• , .� �

a b a bt t E F ii i0 0 1 3( ) = ( ) ¢ = ¢and for• • .� �

aEi


Given any two continuous functions k > 0 and t on an interval I, there exists aunit-speed curve a: I Æ R3 that has these functions as its curvature and torsion.(As we know, any two such curves are congruent.) Thus the natural descrip-tion of curves in R3 is devoid of geometry, consisting of a pair of real-valuedfunctions.

The proof of the existence theorem requires advanced methods, so we havepreferred to illustrate it by the corresponding result for plane curves (Exercises 7–10). Though simpler, this 2-dimensional version has the advan-tage that plane curvature is not required to be positive.

Exercises

1. Given a curve a = (a1, a2, a3): I Æ R3, prove that b: I Æ R3 is con-gruent to a if and only if b can be written as

where ei • ej = dij.

2. Let E1, E2, E3, be a frame field on R3 with dual forms qi and connectionforms wij. Prove that two curves a, b: I Æ R3 are congruent if qi(a ¢) = qi(b¢)and wij(a ¢) = wij(b¢) for 1 � i, j � 3 (Hint: Use Thm. 5.7.)

3. Show that the curve

is a helix by finding its curvature and torsion. Find a helix of the form a(t) = (acos t, asin t, bt) and an isometry F such that F(a) = b.

4. (Computer; see Appendix.) (a) Show that the curves

defined on the entire real line, have the same speed, curvature, and torsion.(b) Find formulas for T and C such that the isometry F = TC carries a to band verify explicitly that F(a) = b. (Hint: Use Ex. 5 of Sec. 2.)

5. (Computer optional.) Is the following curve a helix? Prove your answer.

6. Congruence of curves.(a) Prove that curves a, b: I Æ R2 are congruent if a = b and they havethe same speed.

kk

c t t t t t t t t t t( ) = - + + + + + -( )2 2 2 2 4 2 4cos sin cos sin cos sin ., ,

a bt t t t t t t t t t t t( ) = + - +( ) ( ) = + - -( )2 2 3 2 3 2 31 2 1 2, , , , , ,

b t t t t t t( ) = + -( )3 2 3sin cos sin, ,

b a a at t t t( ) = + ( ) + ( ) + ( )p e e e1 1 2 2 3 3,

k


(b) Show that the space curves

are congruent. Find an isometry that carries a to b.

7. Given a continuous function f on an interval I, prove—using ordinaryintegration of functions—that there exists a unit-speed curve b(s) in R2 forwhich f(s) is the plane curvature. (Hint: Reverse the logic in Ex. 8 of Sec. 2.3.)

8. Show that b(s) = (x(s), y(s)) in the preceding exercise is given by the solu-tions of the differential equations

with initial conditions x(0) = y(0) = j(0) = 0. (These initial conditions suffice,since any other b differs at most by a Euclidean isometry and a reparame-trization s Æ s + c.)

Explicit integration is rarely possible; the following exercises use numeri-cal integration.

9. (Numerical integration, computer graphics.) Write computer commandsthat (a) given f(s), produce a numerical description of the solution curve b(s)in the preceding exercise, and (b) given f(s), plot the solution curve.

10. (Continuation.) Plot unit-speed plane curves with the given plane cur-vature function f on at least the given interval.

(a) f(s) = 1 + es, on -6 £ s £ 3.(b) f(s) = 2 + 3 cos3s, on 0 £ s £ 2 p.(c) f(s) = 3 - 2s2 + s3, on -2.5 £ s £ 3.5.

Adjust scales on axes as needed.

3.6 Summary

The basic result of this chapter is that an arbitrary isometry of Euclideanspace can be uniquely expressed as an orthogonal transformation followedby a translation. A consequence is that the tangent map of an isometry Fis, at every point, essentially just the orthogonal part of F. Then it is a routine matter to test the concepts introduced earlier to see which belong toEuclidean geometry, that is, which are preserved by isometries of Euclideanspace.

¢( ) = ( ) ¢( ) = ( ) ¢( ) = ( )x s s y s s s f scos sin ,j j j, ,

a bt t t t t t t( ) = ( ) ( ) = -( )2 02 2, , and , ,

3.6 Summary 129

Finally, we proved an analogue for curves of the various criteria for con-gruence of triangles in plane geometry; namely, we showed that a necessaryand sufficient condition for two curves in R3 to be congruent is that they havethe same curvature and torsion (and speed). Furthermore, the sufficiencyproof shows how to find the required isometry explicitly.

▼

▲Chapter 4

Calculus on a Surface

130

This chapter begins with the definition of a surface in R3 and with some stan-dard ways to construct surfaces. Although this concept is a more-or-lessfamiliar one, it is not as widely known as it should be that each surface hasa differential and integral calculus strictly comparable with the usual calcu-lus on the Euclidean plane R2. The elements of this calculus—functions,vector fields, differential forms, mappings—belong strictly to the surface andnot to the Euclidean space R3 in which the surface is located. Indeed, we shallsee in the final section that this calculus survives undamaged when R3 isremoved, leaving just the surface and nothing more.

4.1 Surfaces in R3

A surface in R3 is, to begin with, a subset of R3, that is, a certain collectionof points of R3. Of course, not all subsets are surfaces: We must certainlyrequire that a surface be smooth and two-dimensional. These requirementswill be expressed in mathematical terms by the next two definitions.

1.1 Definition A coordinate patch x: D Æ R3 is a one-to-one regularmapping of an open set D of R2 into R3.

The image x(D) of a coordinate patch x—that is, the set of all values ofx—is a smooth two-dimensional subset of R3 (Fig. 4.1). Regularity (Defini-tion 7.9 of Chapter 1), for a patch as for a curve, is a basic smoothness con-dition; the one-to-one requirement is included to prevent x(D) from cuttingacross itself. Initially, in order to avoid certain technical difficulties (Example

4.1 Surfaces in R3 131

1.6), we must use proper patches, those for which the inverse function x-1:x(D) Æ D is continuous (that is, has continuous coordinate functions). If wethink of D as a thin sheet of rubber, then x(D) is gotten by bending andstretching D in a not too violent fashion.

To construct a suitable definition of surface we start from the rough ideathat any small enough region in a surface M resembles a region in the plane R2.The discussion above shows that this can be stated somewhat more preciselyas, near each of its points, M can be expressed as the image of a proper patch.(When the image of a patch x is contained in M, we say that x is a patch inM.) To get the final form of the definition, it remains only to define a neigh-borhood N of p in M to consist of all points of M whose Euclidean distancefrom p is less than some number e > 0.

1.2 Definition A surface in R3 is a subset M of R3 such that for each pointp of M there exists a proper patch in M whose image contains a neighbor-hood of p in M (Fig. 4.2).

The familiar surfaces used in elementary calculus satisfy this definition; forexample, let us verify that the unit sphere S in R3 is a surface. By definition,

FIG. 4.1

FIG. 4.2

132 4. Calculus on a Surface

S consists of all points at unit distance from the origin—that is, all points psuch that

To check the definition above, we start by finding a proper patch in S cover-ing a neighborhood of the north pole (0, 0, 1). Note that by dropping eachpoint (q1, q2, q3) of the northern hemisphere of S onto the xy plane at (q1, q2,0) we get a one-to-one correspondence of this hemisphere with a disk D ofradius 1 in the xy plane (see Fig. 4.3). If this plane is identified with R2 bymeans of the natural association (q1, q2, 0) ´ (q1, q2), then D becomes thedisk in R2 consisting of all points (u, v) such that u2 + v2 < 1. Expressing thiscorrespondence as a function on D yields the formula

Thus x is a one-to-one function from D onto the northern hemisphere ofS. We claim that x is a proper patch. The coordinate functions of x are dif-ferentiable on D, so x is a mapping. To show that x is regular, we computeits Jacobian matrix (or transpose)

where Evidently the rows of this matrix are always linearlyindependent, so its rank at each point is 2. Thus, by the criterion followingDefinition 7.9 of Chapter 1, x is regular and hence is a patch. Furthermore,x is proper, since its inverse function x-1: x(D) Æ D is given by the formula

f u v= - -1 2 2 .

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

Ê

Ë

ÁÁÁ

ˆ

¯

˜˜˜

=

∂∂∂∂

Ê

Ë

ÁÁÁ

ˆ

¯

˜˜˜

uu

vu

fu

uv

vv

fv

fufv

1 0

0 1.

x u v u v u v, , ,( ) = - -( )1 2 2 .

p = + +( ) =p p p12

22

32 1 2

1.

FIG. 4.3


and hence is certainly continuous. Finally, we observe that the patch x coversa neighborhood of (0, 0, 1) in S. Indeed, it covers a neighborhood of everypoint q in the northern hemisphere of S.

In a strictly analogous way, we can find a proper patch covering each ofthe other five coordinate hemispheres of S, and thus verify, by Definition 1.2,that S is a surface. Our real purpose here has been to illustrate Definition1.2—we soon find a much quicker way to prove (in particular) that spheresare surfaces.

The argument above shows that if f is any differentiable real-valued func-tion on an open set D in R2, then the function x: D Æ R3 such that

is a proper patch. We shall call patches of this type Monge patches.We turn now to some standard methods of constructing surfaces. Note that

the image M = x(D) of just one proper patch automatically satisfies 1.2; Mis then called a simple surface. (Thus Definition 1.2 says that any surface inR3 can be constructed by gluing together simple surfaces.)

1.3 Example The surface M: z = f(x, y). Every differentiable real-valuedfunction f on R2 determines a surface M in R3: the graph of f, that is, the setof all points of R3 whose coordinates satisfy the equation z = f(x, y). Evi-dently M is the image of the Monge patch

hence by the remarks above, M is a simple surface.

If g is a real-valued function on R3 and c is a number, denote by M: g = c the set of all points p such that g(p) = c. For example, if g is a temperature distribution in space, then M: g = c consists of all points oftemperature c. There is a simple condition that tells when such a subset ofR3 is a surface.

1.4 Theorem Let g be a differentiable real-valued function on R3, and ca number. The subset M: g(x, y, z) = c of R3 is a surface if the differentialdg is not zero at any point of M.

(In Definition 1.2 and in this theorem we are tacitly assuming that M hassome points in it; thus the equation x2 + y2 + z2 = -1, for example, does notdefine a surface.)

x u v u v f u v, , , ,( ) = ( )( );

x u v u v f u v, , , ,( ) = ( )( )

x- ( ) = ( )11 2 3 1 2p p p p p, , ,


Proof. All we do is give geometric content to a famous result ofadvanced calculus—the implicit function theorem. If p is a point of M, wemust find a proper patch covering a neighborhood of p in M (Fig. 4.4).Since

the hypothesis on dg is equivalent to assuming that at least one of thesepartial derivatives is not zero at p, say (∂g/∂z)(p) π 0. In this case, theimplicit function theorem says that near p the equation g(x, y, z) = c canbe solved for z. More precisely, it asserts that there is a differentiable real-valued function h defined on a neighborhood D of (p1, p2) such that

(1) For each point (u, v) in D, the point (u, v, h(u, v)) lies in M; that is,g(u, v, h(u, v)) = c.

(2) Points of the form (u, v, h(u, v)), with (u, v) in D, fill a neighbor-hood of p in M.

It follows immediately that the Monge patch x: D Æ R3 such that

satisfies the requirements in Definition 1.2. Since p was an arbitrary pointof M, we conclude that M is a surface. ◆

From now on we use the notation M: g = c only when dg π 0 on M. ThenM is a surface said to be defined implicitly by the equation g = c. It is now

x u v u v h u v, , , ,( ) = ( )( )

dggx

dxgy

dygz

dz=∂∂

+∂∂

+∂∂

,

FIG. 4.4


easy to prove that spheres are surfaces. The sphere S in R3 of radius r > 0 and center c = (c1, c2, c3) is the set of all points at distance r from c. If

then S is defined implicitly by the equation g = r2. Now,

Hence dg is zero only at the point c, which is not in S. Thus S is a surface.An important class of surfaces is gotten by rotating curves.

1.5 Example Surfaces of revolution. Let C be a curve in a plane P Ã R3,and let A be a line in P that does not meet C. When this profile curve C isrevolved around the axis A, it sweeps out a surface of revolution M in R3.

Let us check that M really is a surface. For simplicity, suppose that P is acoordinate plane and A is a coordinate axis—say, the xy plane and x axis,respectively. Since C must not meet A, we put it in the upper half, y > 0, ofthe xy plane. As C is revolved, each of its points (q1, q2, 0) gives rise to awhole circle of points

Thus a point p = (p1, p2, p3) is in M if and only if the point

is in C (Fig. 4.5).If the profile curve is C: f(x, y) = c, we define a function g on R3 by

g x y z f x y z, , ,( ) = +( )2 2 .

p = +( )p p p1 22

32 0, ,

q q v q v v1 2 2 0 2, , forcos sin .( ) £ £ p

dg x c dxi i i= -( )Â2 .

g x ci i= -( )Â 2

FIG. 4.5


Then the argument above shows that the resulting surface of revolution isexactly M: g(x, y, z) = c. Using the chain rule, it is not hard to show that dgis never zero on M, so M is a surface.

The circles in M generated under revolution by each point of C are calledthe parallels of M; the different positions of C as it is rotated are called themeridians of M. This terminology derives from the geography of the sphere;however, a sphere is not a surface of revolution as defined above. Its profilecurve must twice meet the axis of revolution, so two “parallels” reduce tosingle points. To simplify the statements of later theorems, we use a slightlydifferent terminology in this case; see Exercise 12.

The necessity of the properness condition on the patches in Definition 1.2is shown by the following example.

1.6 Example Suppose that a rectangular strip of tin is bent into a figure8, as in Fig. 4.6. The configuration M that results does not satisfy our intu-itive picture of what a surface should be, for along the axis A, M is not likethe plane R2 but is instead like two intersecting planes. To express this con-struction in mathematical terms, let D be the rectangle -p < u < p, 0 < v < 1in R2 and define x: D Æ R3 by x(u, v) = (sinu, sin2u, v). It is easy to checkthat x is a patch, but its image M = x(D) is not a surface: x is not a properpatch. Continuity fails for x-1: M Æ D since, roughly speaking, to restore Mto D, x-1 must tear M along the axis A (the z axis of R3).

By Example 1.5, the familiar torus of revolution T is a surface (Fig. 4.16).With somewhat more work, one could construct double toruses of variousshapes, as in Fig. 4.7. By adding “handles” and “tubes” to existing surfacesone can—in principle, at least—construct surfaces of any desired degree ofcomplexity (Fig. 4.8).

FIG. 4.6

FIG. 4.7


Exercises

1. None of the following subsets M of R3 are surfaces. At which points pis it impossible to find a proper patch in M that will cover a neighborhoodof p in M ? (Sketch M—formal proofs not required.)

(a) Cone M: z2 = x2 + y2

(b) Closed disk M: x2 + y2 � 1, z = 0.(c) Folded plane M: xy = 0, x � 0, y � 0.

2. A plane in R3 is a surface M: ax + by + cz = d, where the numbers a, b,c are necessarily not all zero. Prove that every plane in R3 may be describedby a vector equation as on page 62.

3. Sketch the general shape of the surface M: z = ax2 + by2 in each of thefollowing cases:

(a) a > b > 0. (b) a > 0 > b.(c) a > b = 0. (d) a = b = 0.

4. In which of the following cases is the mapping x: R2 Æ R3 a patch?(a) x(u, v) = (u, uv, v). (b) x(u, v) = (u2, u3, v).(c) x(u, v) = (u, u2, v + v3). (d) x(u, v) = (cos2pu, sin2pu, v).

(Recall that x is one-to-one if and only if x(u, v) = x(u1, v1) implies (u, v) =(u1, v1).)

5. (a) Prove that M: (x2 + y2)2 + 3z2 = 1 is a surface.(b) For which values of c is M: z(z - 2) + xy = c a surface?

6. Determine the intersection z = 0 of the monkey saddle

with the xy plane. On which regions of the plane is f > 0? f < 0? How doesthis surface get its name? (Hint: see Fig. 5.19.)

M z f x y f x y y yx: ,= ( ) ( ) = -, , ,3 23

FIG. 4.8

7. Let x: D Æ R3 be a mapping, with

(a) Prove that a point p = (p1, p2, p3) of R3 is in the image x(D) if and onlyif the equations

can be solved for u and v, with (u, v) in D.(b) If for every point p in x(D) these equations have the unique solution

u = f1(p1, p2, p3), v = f2(p1, p2, p3), with (u, v) in D, prove that x is one-to-oneand that x-1: x(D) Æ D is given by the formula

8. Let x: D Æ R3 be the function given by

on the first quadrant D: u > 0, v > 0. Show that x is one-to-one and find aformula for its inverse function x-1: x(D) Æ D. Then prove that x is a properpatch.

9. Let x: R2 Æ R3 be the mapping

Show that x is a proper patch and that the image of x is the entire surfaceM: z = (x2 - y2)/4.

10. If F: R3 Æ R3 is a diffeomorphism and M is a surface in R3, prove that the image F(M) is also a surface in R3. (Hint: If x is a patch in M, thenthe composite function F(x) is regular, since F(x)* = F*x* by Ex. 9 ofSec. 1.7.)

11. Prove this special case of Exercise 10: If F is a diffeomorphism of R3,then the image of the surface M: g = c is : = c, where = g(F -1) and

is a surface. (Hint: If dg(v) π 0 at p in M, show by using Ex. 7 of Sec. 1.7that d (F*v) π 0 at F(p).)

12. Let C be a Curve in the xy plane that is symmetric about the x axis.Assume C crosses the x axis and always does so orthogonally. Explain whythere can be only one or two crossings. Thus C is either an arc or is closed(Fig. 4.9). Revolving C about the x axis gives a surface M, called an aug-mented surface of revolution. Explain how to define patches in M at the crossing points.

gM

ggM

x u v u v u v uv, , ,( ) = + -( ).

x u v u uv v, , ,( ) = ( )2 2

x p p p- ( ) = ( ) ( )( )11 2f f, .

p x u v p x u v p x u v1 1 2 2 3 3= ( ) = ( ) = ( ), , , , ,

x u v x u v x u v x u v, , , , , ,( ) = ( ) ( ) ( )( )1 2 3 .


4.2 Patch Computations 139

4.2 Patch Computations

In Section 1, coordinate patches were used to define a surface; now we con-sider some properties of patches that will be useful in studying surfaces.

Let x: D Æ R3 be a coordinate patch. Holding u or v constant in the func-tion (u, v) Æ x(u, v) produces curves. Explicitly, for each point (u0, v0) in Dthe curve

is called the u-parameter curve, v = v0, of x; and the curve

is the v-parameter curve, u = u0 (Fig. 4.10).Thus, the image x(D) is covered by these two families of curves, which are

the images under x of the horizontal and vertical lines in D, and one curvefrom each family goes through each point of x(D).

2.1 Definition If x: D Æ R3 is a patch, for each point (u0, v0) in D:(1) The velocity vector at u0 of the u-parameter curve, v = v0, is denoted

by xu(u0, v0).

v u vÆ ( )x 0 ,

u u vÆ ( )x , 0

FIG. 4.9

FIG. 4.10


(2) The velocity vector at v0 of the v-parameter curve, u = u0, is denotedby xv(u0, v0).

The vectors xu(u0, v0) and xv(u0, v0) are called the partial velocities of x at(u0, v0) (Fig. 4.11).

Thus xu and xv are functions on D whose values at each point (u0, v0) aretangent vectors to R3 at x(u0, v0). The subscripts u and v are intended tosuggest partial differentiation. Indeed if the patch is given in terms of itsEuclidean coordinate functions by a formula

then it follows from the definition above that the partial velocity functionsare given by

The subscript x (frequently omitted) is a reminder that xu(u, v) and xv(u, v)have point of application x(u, v).

2.2 Example The geographical patch in the sphere. Let S be the sphereof radius r > 0 centered at the origin of R3. Longitude and latitude on theearth suggest a patch in S quite different from the Monge patch used on Sin Section 1. The point x(u, v) of S with longitude u (-p < u < p) and lati-tude v (-p/2 < v < p/2) has Euclidean coordinates (Fig. 4.12).

With the domain D of x defined by these inequalities, the image x(D) of xis all of S except one semicircle from north pole to south pole. The u-

x u v r v u r v u r v, , ,( ) = ( )cos cos cos sin sin .

x

x

u

x

v

x

xu

xu

xu

xv

xv

xv

=∂∂

∂∂

∂∂

ÊËÁ

ˆ¯

=∂∂

∂∂

∂∂

ÊËÁ

ˆ¯

1 2 3

1 2 3

, , ,

, , .

x u v x u v x u v x u v, , , , , , ,( ) = ( ) ( ) ( )( )1 2 3

FIG. 4.11


parameter curve, v = v0, is a circle—the parallel of latitude v0. The v-parameter curve, u = u0, is a semicircle—the meridian of longitude u0.

We compute the partial velocities of x to be

where r denotes a scalar multiplication. Evidently xu always points due east,and xv due north (Fig. 4.13). In a moment we shall give a formal proof thatx is a patch in S.

To test whether a given subset M of R3 is a surface, Definition 1.2 demandsproper patches (and Example 1.6 shows why). But once we know that M isa surface, the properness condition need no longer concern us (Exercise 14

x

x

u

v

u v r v u v u

u v r v u v u v

, , , ,

, , ,

( ) = -( )( ) = - -( )

cos sin cos cos

sin cos sin sin cos .

0

FIG. 4.12

FIG. 4.13

of Section 3). Furthermore, in many situations the one-to-one restriction onpatches can also be dropped.

2.3 Definition A regular mapping x: D Æ R3 whose image lies in asurface M is called a parametrization of the region x(D) in M.

(Thus a patch is merely a one-to-one parametrization.) In favorable casesthis image x(D) may be the whole surface M, and we then have the analogueof the more familiar notion of parametrization of a Curve (see end of Section1.4). Parametrizations will be of first importance in practical computationswith surfaces, so we consider some ways of determining whether a mappingx: D Æ R3 is a parametrization of (part of) a given surface M.

The image of x must, of course, lie in M. Note that if the surface is givenin the implicit form M: g = c, this means that the composite function g(x)must have constant value c.

To test whether x is regular, note first that parameter curves and partialvelocities xu and xv are well-defined for an arbitrary differentiable mappingx: D Æ R3. Also, the last two rows of the cross product

give the (transposed) Jacobian matrix of x at each point. Thus the regular-ity of x is equivalent to the condition that xu ¥ xv is never zero, or, by prop-erties of the cross product, that at each point (u,v) of D the partial velocityvectors of x are linearly independent.

Let us try out these methods on the mapping x given in Example 2.2. Sincethe sphere is defined implicitly by g = x2 + y2 + z2 = r2, we must show thatg(x) = r2. Substituting the coordinate functions of x for x, y, and z gives

A short computation using the formulas for xu and xv, given in Example2.2, yields

r u vU u vU v vUu v- ¥ = + +2 2

12

2 3x x cos cos sin cos cos sin .

r g v u v u v

v v

- ( ) = ( ) + ( ) +

= + =

2 2 2 2

2 2 1

x cos cos cos sin sin

cos sin .

x xu v

U U Uxu

xu

xu

xv

xv

xv

¥ =∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

1 2 3

1 2 3

1 2 3


Since -p/2 < v < p/2 in the domain D of x, cosv is never zero there; but sinuand cosu are never zero simultaneously, so xu ¥ xv, is never zero on D. Thusx is regular—and hence is a parametrization. In fact, it remains a parame-trization if the condition -p < u < p is dropped, thus replacing D by the infi-nite strip -p/2 < v < p/2. In this case the u-parameter curves are periodicparametrizations of the meridians, and x covers the entire sphere except forthe poles (0, 0, ±1).

To show that x on the original domain D is a patch, it remains only toshow that it is one-to-one on D, that is,

In view of the definition of x, the vector equation here gives the three scalarequations

Since -p/2 < v < p/2 in D, the last equation implies v = v1. Thus r cosv =rcosv1 > 0 can be canceled from the first two equations, and we conclude thatu = u1 as well.

The geographical definition of x in Example 2.2 makes the precedingresults seem almost obvious, but the methods used will serve in more diffi-cult cases.

2.4 Example Parametrization of a surface of revolution. Suppose thatM is obtained, as in Example 1.5, by revolving a curve C in the upper halfof the xy plane about the x axis. Now let

be a parametrization of C (note that h > 0). As we observed in Example 1.5,when the point (g(u), h(u), 0) on the profile curve C has been rotated throughan angle v, it reaches a point x(u, v) with the same x coordinate g(u), but newy and z coordinates h(u) cosv and h(u) sinv, respectively (Fig. 4.14). Thus

Evidently this formula defines a mapping into M whose image is all of M. Ashort computation shows that xu and xv are always linearly independent, sox is a parametrization of M. The domain D of x consists of all points (u, v)for which u is in the domain of a. The u-parameter curves of x parametrize

x u v g u h u v h u v, , ,( ) = ( ) ( ) ( )( )cos sin .

a u g u h u( ) = ( ) ( )( ), , 0

r v u r v u

r v u r v u

r v r v

cos cos cos cos

cos sin cos sin

sin sin .

=

=

=

1 1

1 1

1

,

,

x xu v u v u v u v, , , ,( ) = ( ) fi ( ) = ( )1 1 1 1 .



the meridians of M, the v-parameter curves the parallels. (Thus the parame-trization x: D Æ M is never one-to-one.)

Obviously we are not limited to rotating curves in the xy plane about thex axis. But with other choices of coordinates, we maintain the same geomet-ric meaning for the functions g and h: g measures distance along the axis ofrevolution, while h measures distance from the axis of revolution.

Actually, the geographical patch in the sphere is one instance of Example2.4 (with u and v reversed); here is another.

2.5 Example Torus of revolution T. This is the surface of revolutionobtained when the profile curve C is a circle. Suppose that C is the circle in the xz plane with radius r > 0 and center (R, 0, 0). We shall rotate aboutthe z axis; hence we must require R > r to keep C from meeting the axis ofrevolution. A natural parametrization (Fig. 4.15) for C is

Thus by the remarks above we must have g(u) = r sin u (distance along thez axis) and h(u) = R + rcosu (distance from the z axis). The general argument in Example 2.4—with coordinate axes permuted—then yields the parametrization

We call x the usual parametrization of the torus (Fig. 4.16). Its domain is thewhole plane R2, and it is periodic in both u and v:

x x xu v u v u v u v+( ) = +( ) = ( ) ( )2 2p p, , , for all , .

x u v h u v h u v g u

R r u v R r u v r u

, , ,

, ,

( ) = ( ) ( ) ( )( )= +( ) +( )( )

cos sin

cos cos cos sin sin .

a u R r u r u( ) = +( )cos sin .,

FIG. 4.14 FIG. 4.15


2.6 Definition A ruled surface is a surface swept out by a straight line Lmoving along a curve b. The various positions of the generating line Lare called the rulings of the surface. Such a surface always has a ruled parametrization,

We call b the base curve and d the director curve, although d is usually pic-tured as a vector field on b pointing along the line L.

Several examples of ruled surfaces are given in the following exercises. Itis usually necessary to put restrictions on b and d to ensure that x is a parametrization.

There are infinitely many different parametrizations and patches in anysurface. Those we have discussed occur frequently and are fitted in a naturalway to their surfaces.

Exercises

1. Find a parametrization of the entire surface obtained by revolving:(a) C: y = cosh x around the x axis (catenoid).(b) C: (x - 2)2 + y2 = 1 around the y axis (torus).(c) C: z = x2 around the z axis (paraboloid).

2. Partial velocities xu and xv are defined for an arbitrary mapping x: D ÆR3, so we can consider the real-valued functions

on D. Prove

x xu v EG F¥ = -2 2.

E F Gu u u v v v= = =x x x x x x• • •, ,

x u v u v u,( ) = ( ) + ( )b d .

FIG. 4.16


Deduce that x is a regular mapping if and only if EG - F 2 is never zero. (Thisis often the easiest way to check regularity. We will see, beginning in the nextchapter, that the functions E, F, G are fundamental to the geometry ofsurfaces.)

3. A generalized cone is a ruled surface with a parametrization of the form

Thus all rulings pass through the vertex p (Fig. 4.17). Show that x is regularif and only if v and d ¥ d ¢ are never zero. (Thus the vertex is never part ofthe cone. Unless the term generalized is used, we assume that d is a closedcurve and require either v > 0 or v < 0.)

4. A generalized cylinder is a ruled surface for which the rulings are allEuclidean parallel (Fig. 4.18). Thus there is always a parametrization of theform

Prove: (a) Regularity of x is equivalent to b¢ ¥ q never zero.(b) If C: f(x, y) = a is a Curve in the plane, show that in R3 the same equa-tion defines a surface . If t Æ (x(t), y(t)) is a parametrization of C, finda parametrization of that shows it is a generalized cylinder.

Generalized cylinders are a rather broad category—including Euclideanplanes when b is a straight line—so unless the term generalized is used, weassume that cylinders are over closed curves b.

5. A line L is attached orthogonally to an axis A (Fig. 4.19). If L movessteadily along A, rotating at constant speed, then L sweeps out a helicoid H.

When A is the z axis, H is the image of the mapping x: R2 Æ R3 such that

x u v u v u v bv b, , ,( ) = ( ) π( )cos sin .0

CC

x q q Ru v u v,( ) = ( ) + Œ( )b 3 .

x pu v v u,( ) = + ( )d .

FIG. 4.17 FIG. 4.18


(a) Prove that x is a patch.(b) Describe its parameter curves.(c) Express the helicoid in the implicit form g = c.(d) (Computer graphics.) Plot one full turn (0 � v � 2p) of a helicoid withb = 1/2. Restrict the rulings to -1 � u � 1.

6. (a) Show that the saddle surface M: z = xy is doubly ruled: Find tworuled parametrizations with different rulings.

(b) (Computer graphics.) Plot a representative portion of M, using a patchfor which the parameter curves are rulings.

7. Let b be a unit-speed parametrization of the unit circle in the xy plane.Construct a ruled surface as follows: Move a line L along b in such a waythat L is always orthogonal to the radius of the circle and makes constantangle p /4 with b¢ (Fig. 4.20).

(a) Derive this parametrization of the resulting ruled surface M:

(b) Express x explicitly in terms of v and coordinate functions for b.(c) Deduce that M is given implicitly by the equation

(d) Show that if the angle p /4 above is changed to -p /4, the same surfaceM results. Thus M is doubly ruled.(e) Sketch this surface M showing the two rulings through each of thepoints (1, 0, 0) and (2, 1, 2).

x y z2 2 2 1+ - = .

x u v u v u U,( ) = ( ) + ¢( ) +( )b b 3 .

FIG. 4.19 FIG. 4.20


8. Let M be the surface of revolution gotten by revolving the curve t Æ (g(t), h(t), 0) about the x axis (h > 0). Show that:

(a) If g¢ is never zero, then M has a parametrization of the form

(b) If h¢ is never zero, then M has a parametrization of the form

A quadric surface is a surface M: g = 0 in R3 such that g contains at mostquadratic terms in x1, x2, x3, that is,

Trivial cases excepted, every quadric surface is congruent to one of the fivetypes described in the next two exercises. (Use of computers is optional inthese exercises.)

9. In each case, (i) show that M is a surface, and sketch its general shapewhen a = 3, b = 2, c = 1; (ii) show that x is a parametrization in M anddescribe what part of M it covers.

(a) Ellipsoid.

x(u, v) = (a cos u cosv, bcos u sinv, c sinu) on D: -p/2 < u < p/2.

(b) Hyperboloid of one sheet (Fig. 4.21).

x(u, v) = (a cosh ucosv, b cosh u sin v, c sinh u)

on R2.

Mxa

yb

zc

:2

2

2

2

2

21+ - = ,

Mxa

yb

zc

:2

2

2

2

2

21+ + = ,

g a x x b x cij i ji j

i ii

= + +Â Â,

.

x u v f u u v u v, , ,( ) = ( )( )cos sin .

x u v u f u v f u v, , ,( ) = ( ) ( )( )cos sin .

FIG. 4.21

4.3 Differentiable Functions and Tangent Vectors 149

(c) Hyperboloid of two sheets (Fig. 4.21).

x(u, v) = (a sinh u cos v, b sinh u sin v, c cosh u)

on D: u π 0.

10. Sketch the following surfaces (graphs of functions) for a = 2, b = 1:

(a) Elliptic paraboloid. Show that

is a parametrization that omits only one point of M.

(b) Hyperbolic paraboloid.

Show that M is covered by the single patch

11. Doubly ruled quadrics.(a) Show that the hyperbolic paraboloid M in the preceding exercise isdoubly ruled.(b) (Computer graphics.) For a = 2, b = 1 use the patch in (b) of Exercise10 to plot a portion of M. (Keep the same scale on all axes; the parame-ter curves will be the rulings.)(c) Find two different ruled parametrizations of the hyperboloid of onesheet by using the scheme in the special case, Exercise 7.(d) (Computer graphics.) Plot a portion of each of these parametrizations,taking a = 1.5, b = 1, c = 2.

4.3 Differentiable Functions and Tangent Vectors

We now begin an exposition of the calculus on a surface M in R3. The spaceR3 will gradually fade out of the picture, since our ultimate goal is a calcu-lus for M alone. Generally speaking, we shall follow the order of topics inChapter 1, making such changes as are necessary to adapt the calculus of theplane R2 to a surface M.

Suppose that f a is real-valued function defined on a surface M. If x: D ÆM is a coordinate patch in M, then the composite function f(x) is called acoordinate expression for f; it is just an ordinary real-valued function (u, v) Æf(x(u, v)). We define f to be differentiable provided all its coordinate expres-sions are differentiable in the usual Euclidean sense (Definition 1.3 ofChapter 1).

x Ru v a u v b u v uv, , , on( ) = +( ) -( )( )4 2.

M zxa

yb

: .= -2

2

2

2

x u v au v bu v u u, , , , ,( ) = ( ) >cos sin 2 0

M zxa

yb

: .= +2

2

2

2

Mxa

yb

zc

:2

2

2

2

2

21+ - = - ,


For a function F: Rn Æ M, each patch x in M gives a coordinate expres-sion x-1(F ) for F. Evidently this composite function is defined only on the setO of all points p of Rn such that F(p) is in x(D). Again we define F to be dif-ferentiable provided all its coordinate expressions are differentiable in theusual Euclidean sense. We must understand that this includes the requirementthat O be an open set of Rn, so that the differentiability of x-1(F ):O Æ R2 iswell-defined, as in Section 7 of Chapter 1

In particular, a curve a: I Æ M in a surface M is, as before, a differentiablefunction from an open interval I into M.

To see how this definition works out in practice, we examine an importantspecial case.

3.1 Lemma If a is a curve a : I Æ M whose route lies in the image x(D)of a single patch x, then there exist unique differentiable functions a1, a2 onI such that

or in functional notation, a = x(a1, a2). (See Fig. 4.22.)

Proof. By definition, the coordinate expression x-1 a: I Æ D is differen-tiable—it is just a curve in R2 whose route lies in the domain D of x. If a1,a2 are the Euclidean coordinate functions of x-1 a, then

These are the only such functions, for if a = x(b1, b2), then

◆

These functions a1, a2 are called the coordinate functions of the curve a withrespect to the patch x.

a a b b b b1 21 1

1 2 1 2, , ,( ) = = ( ) = ( )- -x x xa .

a a= = ( )-xx x11 2a a, .

a t a t a t t( ) = ( ) ( )( )x 1 2, for all ,

FIG. 4.22


For an arbitrary patch x: D Æ M, it is natural to think of the domain Das a map of the region x(D) in M. The functions x and x-1 establish a one-to-one correspondence between objects in x(D) and objects in D. If a curvea in x(D) represents the voyage of a ship, the coordinate curve (a1, a2) plotsits position on the map D.

A rigorous proof of the following technical fact requires the methods ofadvanced calculus, and we shall not attempt to give a proof here.

3.2 Theorem Let M be a surface in R3. If F: Rn Æ R3 is a (differentiable)mapping whose image lies in M, then considered as a function F. Rn Æ Minto M, F is differentiable (as defined above).

This theorem links the calculus of M tightly to the calculus of R3. Forexample, it implies the “obvious” result that a curve in R3 that lies in M is acurve of M.

Since a patch is a differentiable function from an open set of R2 into R3, itfollows that a patch is a differentiable function into M. Hence its coordinateexpressions are all differentiable, so patches overlap smoothly.

3.3 Corollary If x and y are patches in a surface M in R3 whose imagesoverlap, then the composite functions x-1y and y-1x are (differentiable) map-pings defined on open sets of R2.

FIG. 4.23

The function y-1x, for example, is defined only for those points (u, v) in Dsuch that x(u, v) lies in the image y(E) of y (Fig. 4.23).

By an argument like that for Lemma 3.1, Corollary 3.3 can be rewritten:

3.4 Corollary If x and y are overlapping patches in M, then there existunique differentiable functions and such thatvu


for all (u, v) in the domain of x-1y. In functional notation: y = x( , ).There are, of course, symmetrical equations expressing x in terms of y.Corollary 3.3 makes it much easier to prove differentiability. For example,

if f is a real-valued function on M, instead of verifying that all coordinateexpressions f(x) are Euclidean differentiable, we need only do so for enoughpatches x to cover all of M (so a single patch will often be enough). The proofis an exercise in checking domains of composite functions: For an arbitrarypatch y, fx and x-1y differentiable imply fxx-1y differentiable. This functionis in general not fy, because its domain is too small. But since there areenough x’s to cover M, such functions constitute all of f y, and thus provethat it is differentiable.

It is intuitively clear what it means for a vector to be tangent to a surfaceM in R3. A formal definition can be based on the idea that a curve in M musthave all its velocity vectors tangent to M.

3.5 Definition Let p be a point of a surface M in R3. A tangent vector vto R3 at p is tangent to M at p provided v is a velocity of some curve in M(Fig. 4.24).

The set of all tangent vectors to M at p is called the tangent plane of M atp and is denoted by Tp(M). The following result shows, in particular, that ateach point p of M the tangent plane Tp(M) is actually a 2-dimensional vectorsubspace of the tangent space Tp(R3).

3.6 Lemma Let p be a point of a surface M in R3, and let x be a patchin M such that x(u0, v0) = p. A tangent vector v to R3 at p is tangent to Mif and only if v can be written as a linear combination of xu(u0, v0) and xv(u0, v0).

vu

y xu v u u v v u v, , , ,( ) = ( ) ( )( )

FIG. 4.24

Since partial velocities are always linearly independent, we deduce that theyprovided a basis for the tangent plane of M at each point of x(D).

Proof. Note that the parameter curves of x are curves in M, so xu andxv are always tangent to M at p.

First suppose that v is tangent to M at p; thus there is a curve a in Msuch that a(0) = p and a ¢(0) = v. Now by Lemma 3.1, a may be written

a = x(a1, a2);

hence by the chain rule,

But since a(0) = p = x(u0, v0), we have a1(0) = u0, a2(0) = v0. Hence eval-uation at t = 0 yields

Conversely, suppose that a tangent vector v to R3 can be written

By computations as above, v is the velocity vector at t = 0 of the curve

Thus v is tangent to M at p. ◆

A reasonable deduction, based on the general properties of derivatives, isthat the tangent plane Tp(M) is the linear approximation of the surface Mnear p.

3.7 Definition A Euclidean vector field Z on a surface M in R3 is a func-tion that assigns to each point p of M a tangent vector Z(p) to R3 at p.

A Euclidean vector field V for which each vector V(p) is tangent to M atp is called a tangent vector field on M (Fig. 4.25). Frequently these vectorfields are defined, not on all of M, but only on some region in M. As usual,we always assume differentiability.

A Euclidean vector z at a point p of M is normal to M if it is orthogonalto the tangent plane Tp(M)—that is, to every tangent vector to M at p. Anda Euclidean vector field Z on M is a normal vector field on M provided eachvector Z(p) is normal to M.

t u tc v tcÆ +( )x 0 0 2+ ,1 .

v x x= ( ) + ( )c u v c u vu v1 0 0 2 0 0, , .

v x x= ¢( ) = ( ) ( ) + ( ) ( )a 0 0 010 0

20 0

dadt

u vdadt

u vu v, , .

¢ = ( ) + ( )a x xu va adadt

a adadt1 2

11 2

2, , .



Because Tp(M) is a two-dimensional subspace of Tp(R3), there is only onedirection normal to M at p: All normal vectors z at p are collinear.

Thus if z is not zero, it follows that Tp(M) consists of precisely those vectorsin Tp(R3) that are orthogonal to z.

It is particularly easy to deal with tangent and normal vector fields on asurface given in implicit form.

3.8 Lemma If M: g = c is a surface in R3, then the gradient vector field(considered only at points of M) is a nonvanishing normal

vector field on the entire surface M.

Proof. The gradient is nonvanishing (that is, never zero) on M since inthe implicit case we require that the partial derivatives cannotsimultaneously be zero at any point of M.

We must show that (—g)(p) • v = 0 for every tangent vector v to M at p.First note that if a is a curve in M, then g(a) = g(a1, a2, a3) has constantvalue c. Thus by the chain rule,

Now choose a to have initial velocity

at a(0) = p. Then

◆0 0 0=

∂∂

( )( ) ( ) =∂∂

( ) = —( )( )Â Âgx

ddt

gx

v gi

i

iia

ap p v• .

¢( ) = = ( )a 0 1 2 3v v v v, ,

∂∂

( ) =Â gx

ddti

iaa

0.

∂ ∂g xi

— = Âg g x Ui i∂ ∂

FIG. 4.25


3.9 Example Vector fields on the sphere S: The lemma shows that

is a normal vector field on S (Fig. 4.26). This is geometrically evident,since is the vector p with point of application p! It followsby a remark above that vp is tangent to S if and only if the dot product vp • pp = v • p is zero. Similarly, a vector field V on S is a tangent vector fieldif and only if V • X = 0. For example, V(p) = (-p2, p1, 0) defines a tangentvector field on S that points “due east” and vanishes at the north and southpoles (0, 0, ±r).

We must emphsasize that only the tangent vector fields on M belong to thecalculus of M itself, since they derive ultimately from curves in M (Defini-tion 3.5). This is certainly not the case with normal vector fields. However, aswe shall see in the next chapter, normal vector fields are quite useful in study-ing M from the viewpoint of an observer in R3.

Finally, we shall adapt the notion of directional derivatives to a surface.Definition 3.1 of Chapter 1 uses straight lines in R3; thus we must use the lessrestrictive formulation based on Lemma 4.6 of Chapter 1.

3.10 Definition Let v be a tangent vector to M at p, and let f be a dif-ferentiable real-valued function on M. The derivative v[f] of f with respect to v is the common value of ( fa)(0) for all curves a in M with initialvelocity v.

Directional derivatives on a surface have exactly the same linear and Leib-nizian properties as in the Euclidean case (Theorem 3.3 of Chapter 1).

d dt/( )

X pUi ip p( ) = ( )Â

X g xUi i= — = Â12

g x ri= =Â 2 2.

FIG. 4.26

Exercises

1. Let x be the geographical patch in the sphere S (Ex. 2.2). Find the coordinate expression f(x) for the following functions on S:

(a) f(p) = p12 + p2

2. (b) f(p) = (p1 - p2)2 + p32.

2. Let x be the usual parametrization of the torus (Ex. 2.5).(a) Find the Euclidean coordinates a1, a2, a3 of the curve a(t) = x(t, t).(b) Show that a is periodic, and find its period p > 0, the smallest numbersuch that a(t + p) = a(t) for all t.

3. (a) Prove Corollary 3.4.(b) Derive the chain rule

where xu and xv are evaluated on ( , ).(c) Deduce that yu ¥ yv = Jxu ¥ xv, where J is the Jacobian of the mappingx-1y = ( , ): D Æ R2.

4. Let x be a patch in M.(a) If x* is the tangent map of x (Sec. 7 of Ch. 1), show that

where U1, U2 is the natural frame field on R2.(b) If f is a differentiable function on M, prove

5. Prove that:(a) v = (v1, v2, v3) is tangent to M: z = f(x, y) at a point p of M if and only if

(b) if x is a patch in an arbitrary surface M, then v is tangent to M at x(u, v) if and only if

6. Let x and y be the patches in the unit sphere S that are defined on theunit disk D: u2 + v2 < 1 by

x yu v u v f u v u v v f u v u, , , , , , , , , ,( ) = ( )( ) ( ) = ( )( )

v x x• .u vu v u v, ,( ) ¥ ( ) = 0

vfx

p p vfy

p p v3 1 2 1 1 2 2=∂∂

( ) +∂∂

( ), , .

x x x xu vfu

f fv

f[ ] =∂∂

( )( ) [ ] =∂∂

( )( ), .

x x x x* , * ,U Uu v1 2( ) = ( ) =

vu

vu

y x x y x xu u v v u v

uu

vu

uv

vv

=∂∂

+∂∂

=∂∂

+∂∂

, ,



where (a) On a sketch of S indicate the images x(D) and y(D), and the region onwhich they overlap.(b) At which points of D is y-1x defined? Find a formula for this function.(c) At which points of D is x-1y defined? Find a formula for this function.

7. Find a nonvanishing normal vector field on M: z = xy and two tangentvector fields that are linearly independent at each point.

8. Let C be the circular cone parametrized by

If a is the curve

(a) Express a ¢ in terms of xu and xv.(b) Show that at each point of a, the velocity a ¢ bisects the angle between xu and xv. (Hint: Verify that

,

where xu and xv are evaluated on )(c) Make a sketch of the cone C showing the curve a.

9. If z is a nonzero vector normal to M at p, let be the Euclideanplane through p orthogonal to z. Prove:

(a) If each tangent vector vp to M at p is replaced by its tip p + v, thenTp(M) becomes . Thus gives a concrete representation ofTp(M) in R3. It is called the Euclidean tangent plane to M at p.(b) If x is a patch in M, then consists of all points r in R3 suchthat (r - x(u, v)) • xu(u, v) ¥ xv(u, v) = 0.(c) If M is given implicitly by g = c, then consists of all points rin R3 such that (r - p) • (—g)(p) = 0.

10. In each case below find an equation of the form ax + by + cz = d forthe plane .

(a) p = (0, 0, 0) and M is the sphere

(b) p = (1, -2, 3) and M is the ellipsoid

(c) p = x(2, p /4), where M is the helicoid parametrized by

x u v u v u v v, , ,( ) = ( )cos sin .2

x y z2 2 2

4 16 181+ + = .

x y z2 2 21 1+ + -( ) = .

T Mp ( )

T MP ( )

T Mx u v,( ) ( )

T Mp ( )T Mp ( )

T Mp ( )

2t et,( ).

¢ ÊËÁ

ˆ¯

= ¢ ÊËÁ

ˆ¯

a a• •xx

xx

u

u

v

v

x 2t et,( ):x u v v u u, , ,( ) = ( )cos sin .1

f u v= - -1 2 2 .

11. (Continuation of Ex. 2.) With x the usual parametrization of the torusof revolution T, consider the curve a: R Æ T such that a(t) = x(at, bt).

(a) If a/b is a rational number, show that a is a simple closed curve in T,that is, periodic with no self-crossings.(b) If a/b is irrational, show a is one-to-one. Such a curve is called awinding line on the torus. It is dense in T in the sense that given any e > 0,a comes within distance e of every point of T.(c) (Computer graphics.) For reference, plot the torus T with R = 3, r = 1(see Ex. 2.5). Then plot the following curves in T:(i) a(t) = x(3t, 5t) on intervals 0 � t � b, for b = p, 2p, and larger values.Estimate the period of a, in this case the smallest number T > 0 such thata(T) = a(0).(ii) a(t) = x(pt, 5t) on intervals 0 � t � b, for increasing values of b. (Keepthe curve reasonably smooth.)

12. A Euclidean vector field on M is differentiable provided itscoordinate functions z1, z2, z3 (on M) are differentiable. If V is a tangent vectorfield on M, show that

(a) For every patch x: D Æ M, V can be written as

(b) V is differentiable if and only if the functions f and g (on D) are differentiable.

The following exercises deal with open sets in a surface M in R3, that is, setsU in M that contain a neighborhood in M of each of their points.

13. Prove that if y: E Æ M is a proper patch, then y carries open sets in Eto open sets in M. Deduce that if x: D Æ M is an arbitrary patch, then the image x(D) is an open set in M. (Hint: To prove the latter assertion, useCor. 3.3.)

14. Prove that every patch x: D Æ M in a surface M in R3 is proper. (Hint:Use Ex. 13. Note that (x-1y)y-1 is continuous and agrees with x-1 on an openset in x(D).)

15. If U is a subset of a surface M in R3, prove that U is itself a surface inR3 if and only if U is an open set of M.

4.4 Differential Forms on a Surface

In Chapter 1 we discussed differential forms on R3 only in sufficient detail totake care of the Cartan structural equations (Theorem 8.3 of Chapter 2). Inthe next three sections we shall give a rather complete treatment of forms ona surface.

V u v f u v u v g u v u vu vx x x, , , , ,( )( ) = ( ) ( ) + ( ) ( ).

Z zUi i= Â


Forms are just what we will need to describe the geometry of a surface,but this is only one example of their usefulness. Surfaces and Euclideanspaces are merely special cases of the general notion of manifold (Section 8).Every manifold has a differential and integral calculus—expressed in termsof functions, vector fields, and forms—that generalizes the usual elementarycalculus on the real line. Thus forms are fundamental to all the manybranches of mathematics and its applications that are based on calculus. Inthe special case of a surface, the calculus of forms is rather easy, but it stillgives a remarkably accurate picture of the most general case.

Just as for R3, a 0-form f on a surface M is simply a (differentiable) real-valued function on M, and a 1-form f on M is a real-valued function ontangent vectors to M that is linear at each point (Definition 5.1 of Chapter1). We did not give a precise definition of 2-forms in Chapter 1, but we shalldo so now. A 2-form will be a two-dimensional analogue of a 1-form: a real-valued function, not on single tangent vectors, but on pairs of tangent vectors.(In this context the term “pair” will always imply that the tangent vectorshave the same point of application.)

4.1 Definition A 2-form h on a surface M is a real-valued function onall ordered pairs of tangent vectors v, w to M such that

(1) h (v, w) is linear in v and in w;(2) h (v, w) = -h (w, v).

Since a surface is two-dimensional, all p-forms with p > 2 are zero, by def-inition. This fact considerably simplifies the theory of differential forms ona surface.

At the end of this section we will show that our new definitions are con-sistent with the informal exposition given in Chapter 1, Section 6.

Forms are added in the usual pointwise fashion; we add only forms of thesame degree p = 0, 1, 2. Just as a 1-form f is evaluated on a vector field V,now a 2-form h is evaluated on a pair of vector fields V, W to give a real-valued function h(V, W) on the surface M. Of course, we shall always assumethat the forms we deal with are differentiable—that is, convert differentiablevector fields into differentiable functions.

Note that the alternation rule (2) in Definition 4.1 implies that

for any tangent vector v. This rule also shows that 2-forms are related todeterminants.

h v v,( ) = 0

4.4 Differential Forms on a Surface 159

4.2 Lemma Let h be a 2-form on a surface M, and let v and w be (lin-early independent) tangent vectors at some point of M. Then

Proof. Since h is linear in its first variable, its value on the pair of tangentvectors av + bw, cv + dw is ah(v, cv + dw) + bh(w, cv + dw). Using thelinearity of h in its second variable, we get

Then the alternation rule (2) gives

◆

Thus the values of a 2-form on all pairs of tangent vectors at a point arecompletely determined by its value on any one linearly independent pair. Thisremark is used frequently in later work.

Wherever they appear, differential forms satisfy certain general properties,established (at least partially) in Chapter 1 for forms on R3. To begin with,the wedge product of a p-form and a q-form is always a ( p + q)-form. If p orq is zero, the wedge product is just the usual multiplication by a function. Ona surface, the wedge product is always zero if p + q > 2. So we need a defin-ition only for the case p = q = 1.

4.3 Definition If f and y are 1-forms on a surface M, the wedge productf Ÿ y is the 2-form on M such that

for all pairs v, w of tangent vectors to M.

Note that f Ÿ y really is a 2-form on M, since it is a real-valued functionon all pairs of tangent vectors and satisfies the conditions in Definition 4.1.The wedge product has all the usual algebraic properties except commuta-tivity; in general, if x is a p-form and h is a q-form, then

On a surface the only minus sign occurs in the multiplication of 1-forms,where just as in Chapter 1, we have f Ÿ y = -y Ÿ f, and hence f Ÿ f = 0.

The differential calculus of forms is based on the exterior derivative d. Fora 0-form (function) f on a surface, the exterior derivative is, as before, the

x h h xŸ = -( ) Ÿ1 pq .

f y f y f yŸ( )( ) = ( ) ( ) - ( ) ( )v w v w w v,

h ha b c d ad bcv w v w v w+ +( ) = -( ) ( ), , .

ac ad bc bdh h h hv v v w w v w w, , , ,( ) + ( ) + ( ) + ( ).

h ha b c da b

c dv w v w v w+ , ,+( ) = ( ).


1-form df such that df(v) = v[ f ]. Wherever forms appear, the exterior deriv-ative of a p-form is a (p + 1)-form. Thus, for surfaces the only new defini-tion we need is that of the exterior derivative df of a 1-form f.

4.4 Definition Let f be a 1-form on a surface M. Then the exterior deriv-ative df of f is the 2-form such that for any patch x in M,

As it stands, this is not yet a valid definition; there is a problem of consis-tency. What we have actually defined is a form dxf on the image of each patchx in M. So what we must prove is that on the region where two patchesoverlap, the forms dxf and dyf are equal. Only then will we have obtainedfrom f a single form df on M.

4.5 Lemma Let f be a 1-form on M. If x and y are patches in M, thendxf = dyf on the overlap of x(D) and y(E).

Proof. Because yu and yv are linearly independent at each point, it suf-fices by Lemma 4.2 to show that

Now, as in Corollary 3.4, we write y = x( , ) and deduce by the chainrule that

(1)

where xu and xv are henceforth evaluated on ( , ). Then by Lemma 4.2,

(2)

where J is the Jacobian (∂ /∂u) (∂ /∂v) - (∂ /∂v) (∂ /∂u). Thus it is clearfrom Definition 4.4 that to prove (dyf) (yu, yv) = (dxf) (yu, yv), all we needis the equation.

(3)∂∂

( )( ) -∂∂

( )( ) =∂

∂( )( ) -

∂∂

( )( )ÏÌÓ

¸˝˛u v

Ju vv v v vf f f fy y x x .

vuvu

d J du v u vx xy y x xf f( )( ) = ( )( ), , ,

vu

y x x

y x x

u u v

v u v

uu

vu

uv

vv

=∂∂

+∂∂

=∂∂

+∂∂

,

,

vu

d du v u vy xy y y yf f( )( ) = ( )( ), , .

du vu v v uf f fx x x x,( ) =∂∂

( )( ) -∂∂

( )( ).


It suffices to operate on (f(yv)), for merely reversing u and v willthen yield (f(yu)). Since (3) requires us to subtract these two deriva-tives, we can discard any terms that will cancel when u and v are everywherereversed.

Applying f to the second equation in (1) yields

Hence by the chain rule,

(4)

where in accordance with the remark above we have discarded two sym-metric terms. Next we use the chain rule—and the same remark—to get

(5)

Now reverse u and v in (5) (and also and ) and subtract from (5)itself. The result is precisely equation (3). ◆

It is difficult to exaggerate the importance of the exterior derivative. Wehave already seen in Chapter 1 that it generalizes the familiar notion of dif-ferential of a function, and that it contains the three fundamental derivativeoperations of classical vector analysis (Exercise 8 in Section 1.6). In Chapter2 it is essential to the Cartan structural equations (Theorem 8.3). Perhaps theclearest statement of its meaning will come in Stokes’s theorem (6.5), whichcould actually be used to define the exterior derivative of a 1-form.

On a surface, the exterior derivative of a wedge product displays the samelinear and Leibnizian properties (Theorem 6.4 of Chapter 2) as in R3; seeExercise 3. For practical computations these properties are apt to be moreefficient than a direct appeal to the definition. Examples of this techniqueappear in subsequent exercises.

The most striking property of this notion of derivative is that there are nosecond exterior derivatives: Wherever forms appear, the exterior derivativeapplied twice always gives zero. For a surface, we need only prove this for 0-forms, since even for a 1-form f, the second derivative d(df) is a 3-form, andhence is automatically zero.

4.6 Theorem If f is a real-valued function on M, then d(df) = 0.

Proof. Let y = df, so we must show dy = 0. It suffices by Lemma 4.2to prove that for any patch x in M we have (dy)(xu, xv) = 0. Now usingExercise 4 of Section 3, we get

vu

∂∂

( )( ) =∂∂

( )( ) ∂∂

+ÊËÁ

ˆ¯

∂∂

+∂

∂( )( ) ∂

∂+Ê

ËÁˆ¯

∂∂u v

vu

uv u

uu

vvv u vf f fy x x. . . . . . .

∂∂

( )( ) =∂∂

( )( ) ∂∂

+∂∂

( )( ) ∂∂

+u u

uv u

vvv u vf f fy x x . . . ,

f f fy x xv u v

uv

vv

( ) = ( ) ∂∂

+ ( ) ∂∂

.

∂ ∂/ v( )∂ ∂/ u( )



and similarly

Hence

◆

Many computations and proofs reduce to the problem of showing that twoforms are equal. As we have seen, to do so it is not necessary to check thatthe forms have the same value on all tangent vectors. In particular, if x is acoordinate patch, then

(1) for 1-forms on x(D): f = y if and only if f(xu) = y(xu) and f(xv) =y(xv);

(2) for 2-forms on x(D): m = v if and only if m(xu, xv) = v(xu, xv).

(To prove these criteria, we express arbitrary tangent vectors as linear com-binations of xu and xv.) More generally, xu and xv may be replaced by anytwo vector fields that are linearly independent at each point.

Let us now check that the rigorous results proved in this section are con-sistent with the rules of operation stated in Chapter 1, Section 6.

4.7 Example Differential forms on the plane R2. Let u1 = u and u2 = vbe the natural coordinate functions, and U1, U2 the natural frame field on R2.The differential calculus of forms on R2 is expressed in terms of u1 and u2 asfollows:

If f is a function, f a 1-form, and h a 2-form, then

(1) f = f1du1 + f2du2, where fi = f(Ui).(2) h = gdu1du2, where g = h(U1,U2).(3) for y = g1du1 + g2du2 and f as above,

(4)

(5)

For the proof of these formulas, see Exercise 4.

dfu

fu

du duf f=∂∂

-∂∂

ÊËÁ

ˆ¯

( )2

1

1

21 2 as above .

dffu

dufu

du=∂∂

+∂∂1

12

2.

f yŸ = -( )f g f g du du1 2 2 1 1 2.

du v

fu v

fv uu v v uy y yx x x x

x x,( ) =

∂∂

( )( ) -∂∂

( )( ) =∂ ( )( )

∂ ∂-

∂ ( )( )∂ ∂

=2 2

0.

y x xv vf( ) =

∂∂

( )( ).

y x x x xu u udf fu

f( ) = ( ) = [ ] =∂∂

( )( )

Similar definitions and coordinate expressions may be established on any Euclidean space. In the case of the real line R1, the natural frame fieldreduces to the single vector field U1 for which . All p-formswith p > 0 are zero, and for a 1-form, f = f(U1) dt.

Wherever differential forms are used, the following conditions are funda-mental.

4.8 Definition A differential form f is closed if its exterior derivative iszero, df = 0; and f is exact if it is the exterior derivative of some form,f = dx.

Since d applied twice is always 0, every exact form is closed. In the case ofa surface, since d increases degrees by 1, every 2-form is closed and no 0-form(i.e., function) is exact. Thus 1-forms are the important case, and for a 1-formf exactness always means that there is a function f such that df = f. The ana-lytical and topological consequences of these definitions run deep.

Exercises

1. Prove the Leibnizian formulas

where f and g are functions on M and f is a 1-form.(Hint: By definition, ( ff)(vp) = f(p)f(vp); hence ff evaluated on xu is f(x)f(xu).)

2. (a) Prove formulas (1) and (2) in Example 4.7 using the remark preced-ing Example 4.7. (Hint: Show (du1du2) (U1, U2) = 1.)

(b) Derive the remaining formulas using the properties of d and the wedgeproduct.

3. If f is a real-valued function on a surface, and g is a function on the realline, show that

Deduce that

4. If f, g, and h are functions on a surface M, and f is a 1-form, prove:(a) d( fgh) = ghdf + fhdg + fgdh,(b) d(ff ) = fdf - f Ÿ df, (ff = ff),(c) (df Ÿ dg) (v, w) = v[ f ]w[g] - v[g]w[ f ].

d g f g f df( )( ) = ¢( ) .

v vp pg f g f f( )[ ] = ¢( ) [ ].

d fg g df f dg d f df f d( ) = + ( ) = Ÿ +, ,f f f

U f df dt1 = [ ] =


5. Suppose that M is covered by open sets U1, . . . , Uk, and on each Ui thereis defined a function fi such that fi - fj is constant on the overlap of Ui andUj. Show that there is a 1-form f on M such that f = dfi on each Ui. Gener-alize to the case of 1-forms fi such that fi - fj is closed.

6. Let y: E Æ M be an arbitrary mapping of an open set of R2 into a surfaceM. If f is a 1-form on M, show that the formula

is still valid even when y is not regular or one-to-one.(Hint: In the proof of Lem. 4.5, check that equation (3) is still valid in

this case.)

A patch x in M establishes a one-to-one correspondence between an openset D of R2 and an open set x(D) of M. Although we have emphasized thefunction x: D Æ x(D), there are some advantages to emphasizing instead theinverse function x-1: x(D) Æ D.

7. If x: D Æ M is a patch in M, let and be the coordinate functions ofx-1, so x-1(p) = ( (p), (p)) for all p in x(D). Show that

(a) and are differentiable functions on x(D) such that:

These functions constitute the coordinate system associated with x.

(b) d (xu) = 1, d (xv) = 0,

d (xu) = 0, d (xv) = 1.

(c) If f is a 1-form and h is a 2-form, then

(Hint: for (b) use Ex. 4(b) of Sec. 3.)

8. Identify (or describe) the associated coordinate system , of(a) The polar coordinate patch x(u, v) = (ucosv, usinv) defined on thedomain D: u > 0, 0 < v < 2p.(b) The identity patch x(u, v) = (u, v) in R2.(c) The geographical patch x in the sphere.

vu

f f fh h

= + ( ) = ( ) ( ) = ( )= ( ) = ( )

f du g dv f g

h du dv hu v

u v

˜ ˜ ;

˜ ˜ .

, where ,

, where ,

x x x x

x x x

vv

uu

˜ ˜ .u u v u v u v vx x, , ,( )( ) = ( )( ) =

vuvu

vu

du vu v v uf f fy y y y,( ) =∂∂

( )( ) -∂∂

( )( )



4.5 Mappings of Surfaces

To define differentiability of a function from a surface to a surface, we followthe same general scheme used in Section 3 and require that all its coordinateexpressions be differentiable.

5.1 Definition A function F: M Æ N from one surface to another is differentiable provided that for each patch x in M and y in N the compositefunction y-1Fx is Euclidean differentiable (and defined on an open set of R2).F is then called a mapping of surfaces.

Evidently the function y-1Fx is defined at all points (u, v) of D such thatF(x(u, v)) lies in the image of y (Fig. 4.27). As in Section 3 we deduce fromCorollary 3.3 that in applying this definition, it suffices to check enoughpatches to cover both M and N.

5.2 Example (1) Let S be the unit sphere in R3 (center at 0) but withnorth and south poles removed, and let C be the cylinder based on the unitcircle in the xy plane. So C is in contact with the sphere along the equator.We define a mapping F: S Æ C as follows: If p is a point of S, draw the lineorthogonally out from the z azis through p, and let F(p) be the point at whichthis line first meets C, as in Fig. 4.28. To prove that F is a mapping, we use the geographical patch x in S (Example 2.2), and for C the patch y(u, v) =(cosu, sinu, v). Now x(u, v) = (cosvcosu, cosvsinu, sinv), so from the defi-nition of F we get

But this point of C is y(u, sinv); hence

F u v u u vx , , ,( )( ) = ( )cos sin sin .

FIG. 4.27

4.5 Mappings of Surfaces 167

Applying y-1 to both sides of this equation gives

so y-1Fx is certainly differentiable. (Actually, x does not entirely cover S, butthe missing semicircle can be covered by a patch like x.) We conclude that Fis a mapping.

(2) Stereographic projection of the punctured sphere S onto the plane. LetS be a unit sphere resting on the xy plane at the origin, so the center of S isat (0, 0, 1). Delete the north pole n = (0, 0, 2) from S. Now imagine that thereis a light source at the north pole, and for each point p of S, let P(p) be theshadow of p in the xy plane (Fig. 4.29). As usual, we identify the xy planewith R2 by (p1, p2, 0) ´ (p1, p2). Thus we have defined a function P from Sonto R2. Evidently P has the form

where r and R are the distances from p and P(p), respectively, to the z axis.But from the similar triangles in Fig. 4.30, we see that R/2 = r/(2 - p3); hence

Now if x is any patch in S, the composite function Px is Euclidean differ-entiable, so P: S Æ R2 is a mapping.

P p p pp

pp

p1 2 31

3

2

3

22

22

, , ,( ) =- -

ÊËÁ

ˆ¯.

P p p pRp

rRp

r1 2 31 2, , , ,( ) = Ê

Ëˆ¯

y x-( )( ) = ( )1F u v u v, , ,sin

F u v u vx y, ,( )( ) = ( )sin .

FIG. 4.28 FIG. 4.29


Just as for mappings of Euclidean space, each mapping of surfaces has atangent map.

5.3 Definition Let F: M Æ N be a mapping of surfaces. The tangent mapF* of F assigns to each tangent vector v to M the tangent vector F*(v) to Nsuch that if v is the initial velocity of a curve a in M, then F*(v) is the initialvelocity of the image curve F(a) in N (Fig. 4.31).

Furthermore, at each point p, the tangent map F* is a linear transforma-tion from the tangent plane Tp(M) to the tangent plane TF(p)(N) (see Exercise9). It follows immediately from the definition that F* preserves velocities ofcurves: If = F(a) is the image in N of a curve a in M, then F*(a ¢) = ¢.As in the Euclidean case, we deduce the convenient property that the tangentmap of a composition is the composition of the tangent maps (Exercise 10).

The tangent map of a mapping F: M Æ N may be computed in terms ofpartial velocities as follows. If x: D Æ M is a parametrization in M, let y bethe composite mapping F(x): D Æ N (which need not be a parametrization).Obviously, F carries the parameter curves of x to the corresponding para-meter curves of y. Since F* preserves velocities of curves, it follows at oncethat

Since xu and xv give a basis for the tangent space of M at each point of x(D),these readily computable formulas completely determine F*.

F Fu u v v* , *x y x y( ) = ( ) = .

aa

FIG. 4.30

FIG. 4.31

The discussion of regular mappings in Section 7 of Chapter 1 translateseasily to the case of a mapping of surfaces F: M Æ N. F is regular providedall of its derivative maps F*p: Tp(M) Æ TF(p)(N) are one-to-one. Since thesetangent planes have the same dimension, we know from linear algebra thatthe one-to-one requirement is equivalent to F* being a linear isomorphism.A mapping F: M Æ N that has an inverse mapping F-1: N Æ M is called adiffeomorphism. We may think of a diffeomorphism F as smoothly distort-ing M to produce N. By applying the Euclidean formulation of the inversefunction theorem to a coordinate expression y-1Fx for F, we can deduce thisextension of the inverse function theorem (7.10 of Chapter 1).

5.4 Theorem Let F: M Æ N be a mapping of surfaces, and suppose thatF*p: Tp(M) Æ TF(p)(N) is a linear isomorphism at some one point p of M.Then there exists a neighborhood U of p in M such that the restriction of Fto U is a diffeomorphism onto a neighborhood V of F(p) in N.

An immediate consequence is this useful result: A one-to-one regularmapping F of M onto N is a diffeomorphism. For since F is one-to-one andonto, it has a unique inverse function F -1, and F -1 is a differentiable mappingsince on each neighborhood V as above, it coincides with the inverse of thediffeomorphism U Æ V. Surfaces M and N are said to be diffeomorphic ifthere exists a diffeomorphism from M to N.

Diffeomorphisms have little respect for size or shape; here are some examples.

5.5 Example (1) Any open rectangle in the plane R2 is diffeomorphic to the entire plane. Take R: -p/2 < u, v < p/2 for simplicity. Then F(u,v) =(tanu, tanv) is a mapping of R onto R2. Using a branch of the inverse tangent function, the mapping F -1(u1, v1) = (tan-1 (u1), tan-1 (v1)) is a dif-ferentiable inverse of F, so F is a diffeomorphism.

(2) The sphere S minus one point is also diffeomorphic to the entire plane.Stereographic projection P, as in Example 5.2(2), is a one-to-one mapping ofthe punctured sphere S0 onto R2. A variant

of the usual geographical parametrization is a parametrization of S0 - 0. Theformula for P in Example 5.2 gives

y xu v P u vvv

u u, , ,( ) = ( )( ) =-

( )21

cossin

cos sin .

x u v v u v u v, , ,( ) = +( )cos cos cos sin sin1



Since P*(xu) = yu and P*(xv) = yv, the regularity of F can be checked bycomputing yu and yv. These turn out to be orthogonal and nonzero, hencelinearly independent, as suggested by Fig. 4.32.

(3) A cylinder C over a closed curve is diffeomorphic to the plane minus one point. For simplicity, take C: x2 + y2 = 1, and define a mapping F: C Æ R2 by F(x, y, z) = ez(x, y). Since ez takes on all values r > 0, F maps C onto R2 - 0.

For the inverse of F, experimentation suggests

To prove G = F -1, compute G(F(x, y, z)) = (x, y, z) and F(G(u, v)) = (u, v).

Differential forms have the remarkable property that they can be movedfrom one surface to another by means of an arbitrary mapping.† Let usexperiment with a 0-form, that is, a real-valued function. If F: M Æ N is amapping of surfaces and f is a function on M, there is simply no reasonablegeneral way to move f over to a function on N. But if instead, f is a functionon N, the problem is easy; we pull f back to the composite function f(F) onM. The corresponding pull-back for 1-forms and 2-forms is accomplished asfollows.

5.6 Definition Let F: M Æ N be a mapping of surfaces.(1) If f is a 1-form on N, let F*f be the 1-form on M such that

for all tangent vectors v to M.(2) If h is a 2-form on N, let F*h be the 2-form on M such that

F F* *f f( )( ) = ( )v v

G u vu

u v

v

u vu v, , ,( ) =

+ ++Ê

Ëˆ¯2 2 2 2

2 2log .

FIG. 4.32

† This is not true for vector fields.


for all pairs of tangent vectors v, w on M (Fig. 4.33).

When we are dealing with a function f in its role as a 0-form, we shall some-times write F*f instead of f(F), in accordance with the notation for the pull-back of 1-forms and 2-forms.

The essential operations on forms are sum, wedge product, and exteriorderivative; all are preserved by mappings.

5.7 Theorem Let F: M Æ N be a mapping of surfaces, and let x and hbe forms on N. Then

(1) F*(x + h) = F*x + F*h,(2) F*(x Ÿ h) = F*x Ÿ F*h,(3) F*(dx) = d(F*x).

Proof. In (1), x and h are both assumed to be p-forms (degree p = 0, 1,2) and the proof is a routine computation. In (2), we must allow x and hto have different degrees. When, say, x is a function f, the given formulameans simply F *( fh) = f(F )F *(h). In any case, the proof of (2) is also astraightforward computation. But (3) is more interesting. The easier casewhen x is a function is left as an exercise, and we address ourselves to thecase where x is a 1-form.

It suffices to show that for every patch x: D Æ M,

Let y = F(x), and recall that F*(xu) = yu and F*(xv) = yv. Thus, using thedefinitions of d and F*, we get

d F F du v u v* , * ,x x( )( )( ) = ( )( )( )x x x x .

F F F* , * , *h h( )( ) = ( )v w v w

FIG. 4.33


Even if y is not a patch, Exercise 6 of Section 4 shows that this last expres-sion is still equal to dx(yu, yv). But

Thus we conclude that d(F*x) and F*(dx) have the same value on (xu, xv). ◆

The elegant formulas in Theorem 5.7 are the key to the deeper study ofmappings. In Chapter 6 we shall apply them to the connection forms of framefields to get fundamental information about the geometry of mappings ofsurfaces.

Exercises

1. Let M and N be surfaces in R3. If F: R3 Æ R3 is a mapping such that theimage F(M) of M is contained in N, then the restriction of F to M is a function F |M: M Æ N. Prove that F |M is a mapping of surfaces. (Hint: UseThm. 3.2.)

2. Let S be the sphere of radius r with center at the origin of R3. Describethe effect of the following mappings F: S Æ S on the meridians and paral-lels of S.

(a) F(p) = -p.(b) F(p1, p2, p3) = (p3, p1, p2).

(c)

3. Let M be a simple surface, that is, one that is the image of a single properpatch x: D Æ R3. If y: D Æ N is any mapping into a surface N, show thatthe function F: M Æ N such that

is a mapping of surfaces. (Hint: Write F = yx-1, and use Cor. 3.3.)

F u v u v u v Dx y, , for all , in( )( ) = ( ) ( )

F p p pp p p p

p1 2 31 2 1 2

32 2

, , , ,( ) =+ -

-ÊË

ˆ¯.

d d F F F du v u v u vx x xy y x x x x, * , * * ,( ) = ( ) = ( )( )( ).

d Fu

Fv

F

uF

vF

u v

u v v u

v u

v u

* , * *

* *

x x x

x x

x x

( )( ) =∂∂

( )( ){ } -∂∂

( )( ){ }

=∂∂

( ){ } -∂∂

( ){ }

=∂∂

( ){ } -∂∂

( ){ }

x x x x

x x

y y .

4. Use the preceding exercise to construct a mapping of the helicoid H (Ex.2.5) onto the torus T (Ex. 2.5) such that the rulings of H are carried to themeridians of T.

5. If S is the sphere || p || = r, the mapping A: S Æ S such that A(p) = -pis called the antipodal map of S. Prove that A is a diffeomorphism and thatA*(vp) = (-v)-p.

6. A regular mapping F: M Æ N of surfaces is often called a local diffeo-morphism. For such a mapping F, prove that, in fact, every point p of M hasa neighborhood U such that F |U is a diffeomorphism of U onto a neighbor-hood of F(p) in N.

7. If x: D Æ M is a parametrization, prove that the restriction of x to asufficiently small neighborhood of a point (u0, v0) in D is a patch in M. (Thusany parametrization can be cut into patches.)

8. Let F: M Æ N be a mapping. If x is a patch in M, then as in the text,let y = F(x). (Although y maps into N, it is not necessarily a patch.) For acurve

in M, show that the image curve = F(a) in N, has velocity

9. Prove: (a) The invariance property needed to justify the definition (5.3)of the tangent map.

(b) Tangent maps F*: Tp(M) Æ TF(p)(N) are linear transformations.

10. Given mappings be the compositemapping. Show that

(a) GF is differentiable, (b) (GF)* = G*F*,(c) (GF)* = F*G*,that is, for any form x on P, (GF )* (x) = F* (G*(x)). (Note the reversal offactors, caused by the fact that forms travel in the opposite direction frompoints and tangent vectors.)

11. Prove that every surface of revolution is diffeomorphic to either a torusor a cylinder. (Hint: Parametrize profile curves on the same interval.) (As Fig. 4.9 suggests, every augmented surface of revolution is diffeomorphic toeither a plane or a sphere.)

12. (a) Show that the inverse mapping P-1 of the stereographic projectionP: S0 Æ R2 is given by

F GM N P GF M PÆ Æ Æ, :let

¢ = ( ) + ( )adadt

a adadt

a au v1

1 22

1 2y y, , .

a

a t a t a t( ) = ( ) ( )( )x 1 2,



(Check that both PP -1 and P -1P are identity maps.)(b) Deduce that the entire sphere S can be covered by only two patches.(The scheme in Section 1 requires six.)

13. (Consistent formulas.) If G: Æ M is a regular mapping onto M, and: Æ N is an arbitrary mapping, we say that the formula F(G(q)) = (q)

is consistent provided

for q1, q2 in . Prove:(a) In this case, F is a well-defined differentiable mapping from M to N.(b) Furthermore, if the reverse implication

also holds, then F is one-to-one.This result is helpful in constructing maps F: M Æ N with specified prop-

erties. Often G will be a parametrization of M.

˜ ˜F F G Gq q q q1 2 1 2( ) = ( ) fi ( ) = ( )

M

G G F Fq q q q1 2 1 2( ) = ( ) fi ( ) = ( )˜ ˜

FMFM

P u vu v f

ff u v- ( ) =

( )+

= +1 2 24 4 24

,, ,

, where .

M

M

F

FN

G˜

˜

4.6 Integration of Forms

Differential forms are no less important in integral calculus than in differen-tial calculus. Indeed, they are just what is needed to establish integrationtheory on an arbitrary surface. In a sense, integration takes place only onEuclidean space, so a form on a surface is integrated by first pulling it backto Euclidean space.

Consider the one-dimensional case. Let a: [a, b] Æ M be a curve segmenton a surface M. The pullback a*f of a 1-form f on M to the interval [a, b]has the expression f(t)dt, where by the remarks following Example 4.7,

Thus the scheme mentioned above yields the following result:

f t U t U t t( ) = ( ) ( )( ) = ( )( )( ) = ¢( )( )a f f a f a* 1 1* .

4.6 Integration of Forms 175

6.1 Definition Let f be a 1-form on M, and let a: [a, b] Æ M be a curvesegment in M (Fig. 4.34). Then the integral of f over a is

The integral often called a line integral, has a wide variety of uses in science and engineering. For example, let us consider a vector field V on asurface M as a force field, and a curve a: [a, b] Æ M as a description of amoving particle, with a(t) its position at time t. What is the total amount of work W done by the force on the particle as it moves from p = a(a) to q = a(b)? The discussion of velocity in Chapter 1, Section 4, shows that forDt small, the subsegment of a from a(t) to a(t + Dt) is approximated by thestraight line segment Dt a(t). Work is done on the particle only by the com-ponent of force tangent to a, that is,

where J is the angle between V(a(t)) and a ¢(t) (Fig. 4.35). Thus the workdone by the force during time Dt is approximately the force (as above) timesthe distance ||a ¢(t)|| Dt. Adding these contributions over the whole time inter-val [a, b] and taking the usual limit, we get

W V t t dta

b

= ( )( ) ¢( )Ú a a• .

V Vaaa

a J( ) ¢¢

= ( )• cos ,

faÚ ,

f a f f aa a b a

b

t dtÚ Ú= = ¢( )( )[ ]Ú *

,.

FIG. 4.34

FIG. 4.35


To express this more simply, we introduce the 1-form f dual to the vector fieldV; its value on a tangent vector w at p is w • V(p). Then by Definition 6.1,the total work is just

We emphasize that this notion of line integral—like everything we do withforms—applies without essential change if the surface M is replaced by aEuclidean space or, indeed, by any manifold (Section 8).

When the 1-form being integrated is the differential of a function, we getthe following generalization of the fundamental theorem of calculus.

6.2 Theorem Let f be a function on M, and let a: [a, b] Æ M be a curvesegment in M from p = a(a) to q = a(b). Then

Proof. By definition,

But

Hence, by the fundamental theorem of calculus,

◆

Thus the integral is path independent: its value is the same for all curves

from p to q. Hence it is zero for all closed curves, a(a) = a(b).The preceding theorem can be interpreted roughly as follows: The “bound-

ary” of the curve segment a from p to q is q - p, where the purely formal minussign indicates that p is the starting point of a. Then the integral of df over aequals the “integral” of f over the boundary of a, namely, f(q) - f(p). Thisinterpretation will be justified by the analogous theorem (6.5) in dimension 2.

If we consider a closed rectangle R: a � u � b, c � v � d in R2 as a 2-dimensional interval, then a 2-segment is a differentiable map x: R Æ M ofR into M (Fig. 4.36). As for a 1-segment, differentiability means that x canbe extended over a larger open set containing R. Although we use the patchnotation x, we do not assume that x is either regular or one-to-one—but thepartial velocities xu and xv are still well defined.

dfaÚ

dfddt

f dt f b f a f fa a

b

Ú Ú= ( )( ) = ( )( ) - ( )( ) = ( ) - ( )a a a q p .

df fddt

f¢( ) = ¢[ ] = ( )( )a a a .

df df dta

b

aaÚ Ú= ¢( ) .

df f faÚ = ( ) - ( )q p .

Wa

= Ú f.


If h is a 2-form on M, then the pullback x*h of h has, using Example 4.7,the expression h du dv, where

Then strict analogy with Definition 6.1 yields:

6.3 Definition Let h be a 2-form on M, and let x: R Æ M be a 2-segmentin M. The integral of h over x is

The physical applications of this integral are no less rich than those of Def-inition 6.1; however, we proceed directly toward the analogue of Theorem6.2.

6.4 Definition Let x: R Æ M be a 2-segment in M, with R the closed rec-tangle a � u � b, c � v � d (Fig. 4.37). The edge curves of x are the curvesegments a, b, g, d such that

Then the boundary ∂x of the 2-segment x is the formal expression

∂ = + - -x a b g d .

a

b

g

d

u u c

v b v

u u d

v a v

( ) = ( )( ) = ( )( ) = ( )( ) = ( )

x

x

x

x

, ,

, ,

, ,

, .

h h hx

x xÚÚ ÚÚ Ú= = ( )Úx du dvR

u vc

d

a

b

* , .

h U U U U u v= ( )( ) = ( ) = ( )x x x x x* , * , * ,h h h1 2 1 2 .

FIG. 4.36

FIG. 4.37


These four curve segments are gotten by restricting the function x: R Æ Mto the four line segments that make up the boundary of the rectangle R (Fig.4.37). The formal minus signs before g and d signal that the parametrizationsof g and d would have to be reversed to give a consistent trip around the rimof x(R) (Fig. 4.38).

Then if f is a 1-form on M, the integral of f over the boundary ∂x of x isdefined to be

The 2-dimensional analogue of Theorem 6.2 is

6.5 Theorem (Stokes’ theorem) If f is a 1-form on M, and x: R Æ M isa 2-segment, then

Proof. We work on the double integral and show that it turns into theintegral of f over the boundary of x. Combining Definitions 6.3 and 4.4gives

Let f = f(xu) and g = f(xv). Then the equation above becomes

(1)

Now we treat these double integrals as iterated integrals. Suppose therectangle R is given, as usual, by the inequalities a � u � b, c � v � d.Then integrating first with respect to u, we find

∂∂

∂∂

gudu dv I v dv I v

gu

u v duR c

b

a

b

ÚÚ Ú Ú= ( ) ( ) = ( ), where , .

dgudu dv

fv

du dvR R

f∂∂

∂∂xÚÚ ÚÚ ÚÚ= - .

d du dvu

du dvvR

u v v uR

f f fÚÚ ÚÚ( ) =∂

∂( ) - ( )( )( )Ê

ËÁˆ¯

∂∂

x x x x, .

df f∂x xÚÚ Ú= .

f f f f f∂ a b g dxÚ Ú Ú Ú Ú= + - - .

FIG. 4.38


In the partial integral defining I(v), v is constant, so the integrand is justthe ordinary derivative with respect to u. Thus, the fundamental theoremof calculus applies to give I(v) = g(b, v) - g(a, v) (Fig. 4.39). Hence

(2)

Consider the first integral on the right. By definition, g(b, v) = f(xv(b,v)). But xv(b, v) is exactly the velocity b¢(v) of the “right side” curve b in∂x. Hence by Definition 6.1,

A similar argument shows that the other integral in (2) is . Thus

(3)

In the same way—but integrating first with respect to v—we find

(4)

Assembling the information in (1), (3), and (4) gives the required result,

◆

Stokes’ theorem ranks as one of the most useful results in all mathe-matics. Alternative formulations and extensive applications can be found intexts on advanced calculus and applied mathematics. We will use it to studythe geometry and topology of surfaces.

The line integral is not particularly sensitive to reparametrization of

the curve segment a. All that matters is the overall direction in which theroute of a is traversed, as indicated by what the reparametrization does toend points.

faÚ

df f f f f fb d g ax xÚÚ Ú Ú Ú Ú Ú= -Ê

Ëˆ¯ - -Ê

Ëˆ¯ =

∂.

∂∂

f fg a

fv

du dvRÚÚ Ú Ú= - .

∂∂

f fb d

gu

du dvRÚÚ Ú Ú= - .

fdÚ

g b v dv v dvc

d

c

d

, .( ) = ¢( )( ) =Ú Ú Úf b fb

∂∂

gudu dv g b v dv g a v dv

R c

d

c

d

ÚÚ Ú Ú= ( ) - ( ), , .

FIG. 4.39

6.6 Lemma Let a(h): [a, b] Æ M be a reparametrization of a curvesegment a: [c, d] Æ M by h: [a, b] Æ [c, d]. For any 1-form f on M,

(1) If h is orientation-preserving, that is, if h(a) = c and h(b) = d, then

(2) If h is orientation-reversing, that is, if h(a) = d and h(b) = c, then

Proof. The velocity of a(h) is a(h)¢ = dh/dt a ¢(h), so

Now we apply the theorem on change of variables in an integral to the lastintegral above. If h is orientation-preserving, then

while in the orientation-reversing case,

◆

This lemma provides a concrete interpretation to the formal minus signsin the boundary ∂x = a + b - g - d of a 2-segment. For any curve

x: [t0, t1] Æ M,

let -x be any orientation-reversing reparametrization of x, for instance,(-x)(t) = x(t0 + t1 - t). Then by the lemma,

Thus the formula for just before Theorem 6.5 can be rewritten as

Exercises

1. If a is a curve in R2 and f is a 1-form, prove this computational rule forfinding f(a ¢)dt: Substitute u = a1 and v = a2 into the coordinate expression f = f(u, v) du + g(u, v) dv.

2. Let a: [-1, 1] Æ R2 be the curve segment given by a(t) = (t, t2).

(a) If f = v2 du + 2uv dv, compute

(b) Find a function f such that df = f and check Theorem 6.2 in this case.

faÚ .

f f f f fa b g d∂ - -Ú Ú Ú Ú Ú= + + +

x.

f∂Ú x

f fx x-Ú Ú= - .

f f a f a fa ah d

c

c

d

du du( )Ú Ú Ú Ú= ¢( ) = - ¢( ) = - .

f f a fa ah c

d

du( )Ú Ú Ú= ¢( ) = ,

f f a f aa h a

b

a

b

h dt hdhdt

dt( )Ú Ú Ú= ( )¢Ê

Ëˆ¯ = ¢( )( ) .

f fa ah( )Ú Ú= - .

f fa ah( )Ú Ú= .


3. Let f be a 1-form on a surface M. Show:(a) If f is closed, then for every 2-segment x in M.(b) If f is exact, then more generally,

for any closed, piecewise smooth curve whose smooth segments are a1, . . . ,ak (hence ak ends at the start of a1).

4. The 1-form

is well-defined on the plane R2 with the origin 0 removed. Show:(a) y is closed but not exact on R2 - 0. (Hint: Integrate around the unitcircle and use Ex. 3.)(b) The restriction of y to, say, the right half-plane u > 0 is exact.

5. (a) Show that every curve a in R2 that does not pass through the originhas an (orientation-preserving) reparametrization in the polar form

(Hint: Use Ex. 12 of Sec. 2.1.)If the curve a: [a, b] Æ R2 - 0 is closed, prove:

(b) wind is an integer.

This integer, called the winding number of a about 0, represents the totalalgebraic number of times a has gone around the origin in the counter-clockwise direction. (Note that wind(a) = wind(a/||a||).)

(c) If y is the 1-form in Exercise 4, then wind

(d) If a = ( f, g), then

(The determinant is of the 2¥2 matrix whose rows are a(t) and a ¢(t).)

6. (Continuation, by computer.) For a point p Œ R2 not on a closed curve a,the winding number of a about p is defined to be wind(a - p).

(a) Write commands that given a closed curve a in R2 and a point p noton a, return the winding number of a about p. (Hint: Use either integralin (d) of the preceding exercise.)

wind,

ap p

a aa a

( ) =¢ - ¢

+=

( ) ¢( )( )( ) ( )Ú Ú1

21

22 2

fg gff g

dtt t

t tdt

a

b

a

b det•

.

ap

ya

( ) = Ú12

.

ap

( ) =( ) - ( )J Jb a

2

a J Jt r t t t tr( ) = ( ) ( ) ( ) ( )( )cos .sin,

y =-+

u dv v duu v2 2

f fa aÚ Ú= =Â

ii

0

f∂Ú x

= 0


(b) In each case, plot the curve a (observing its orientation) and estimatethe winding numbers about the indicated points p. Then calculate thesenumbers, using the integral in (d) above. (Numerical integration is efficienthere, since the result is known to be an integer.)

(i) lemniscate, a(t) = (2sin t, sin2t); p = (1, 0), (0, 1), (-1, 0).(ii) limaçon, b(t) = (3sin t + 1) (cos t, sin t); p = (0, 1), (0, 3), (0, 5).

7. Let F: M Æ N be a mapping. Prove:(a) If a is a curve segment in M, and f is a 1-form on N, then

(b) If x is 2-segment in M, and h is a 2-form on N, then

8. Let x be a patch in a surface M. For a curve segment

in x(R), show that

where xu and xv are evaluated on (a1, a2). (This generalizes Ex. 1, which isrecovered by using the identity patch x(u, v) = (u, v) in R2.)

9. Let x be the usual parametrization of the torus T (Ex. 2.5). For integersm and n, let a be the closed curve

Find:

(a) where x is the 1-form such that x(xu) = 1 and x(xv) = 0.

(b) where h is the 1-form such that h(xu) = 0 and h(xv) = 1.

For an arbitrary closed curve g in T, is an integer that counts the

total (algebraic) number of times g goes around the torus in the general

direction of the parallels, and gives a similar count for the meridi-

ans. (This suggests the informal notation x = dJ, h = dj, but see Ex. 7 ofSec. 7.)

10. Let x: R Æ M be a 2-segment defined on the unit square R: 0 � u,v � 1. If f is the 1-form on M such that

f fx xu vu v uv( ) = + ( ) =and ,

h pg

2( )Ú

x pg

2( )Úh

aÚ ,

xaÚ ,

a pt mt nt t( ) = ( ) ( )x , 0 2� � .

f f faÚ Ú= ( ) + ( )Ê

ËÁˆ¯

x xu va

b dadt

dadt

dt1 2 ,

a t a t a t a t b( ) = ( ) ( )( )x 1 2, , ,� �

FF

* .h hx xÚÚ ÚÚ=

( )

FF

*f fa aÚ Ú=

( ).



compute and separately, and check the results by Stokes’ theorem.

(Hint: x*df = d(x*f).)

11. Same as Exercise 10, except that R: 0 � u � p/2, 0 � v � p, and f(xu) = ucosv, f(xv) = vsinu.The following exercise is a 2-dimensional analogue of Lemma 6.6. However,with future applications in mind, we generalize 2-segments x: R Æ M byreplacing the rectangle R by any compact region R in R2 whose boundaryconsists of smooth curve segments. (Compactness ensures that integrals overR will be finite.)

12. (Effect of change of variables.) Let x: S Æ M be a differentiable mappingand let (U, V): R Æ S be a one-to-one regular map whose Jacobian determinant

is always positive (orientation-preserving case) or always negative (orientation-reversing case). Then let y: R Æ M be given by y(u, v) = x(U(u, v), V(u, v)).

(a) For a 2-form h on M, use

to prove

(b) Deduce that in the orientation-preserving case, and

minus this in the orientation-reversing case.(Hint: The formula for change of variables in a double integral involvesthe absolute value of a Jacobian determinant.)

13. The classical Stokes’ theorem asserts, typically, that if x: D Æ R3 is a 2-segment and V is a vector field on R3, then

where — ¥ V = curl V. Interpret this as a special case of Theorem 6.5. (Hint:Assume and use Ex. 8 of Sec. 1.6.)dA EG F du dvª - 2 † ,

V ds U V dA• • ,∂Ú ÚÚ= — ¥( )x x

h hx yÚÚ ÚÚ=

h hx x y yu v U VJ u v U V U V, , , , ,( ) = ( ) ( ) ( )( ).

x y y x y yu U V v U V

Uu

Vu

Uv

Vv

=∂∂

+∂∂

=∂∂

+∂∂

,

J u v

Uu

Uv

Vu

Vv

,( ) =

∂∂

∂∂

∂∂

∂∂

È

Î

ÍÍÍÍ

˘

˚

˙˙˙˙

det

f∂Ú x

dfxÚÚ

† This geometric result will be clarified later on. Note that Theorem 6.5 does not involve geometry.

4.7 Topological Properties of Surfaces

Topological properties are the most basic a surface can have. In this sectionwe discuss four such properties, phrasing the definitions in terms most effi-cient for geometry.

7.1 Definition A surface is connected provided that for any two points pand q of M there is a curve segment in M from p to q. (See also Exercise 9.)

Thus a connected surface M is all in one piece, since one can travel fromany point in M to any other without leaving M. Most of the surfaces we havemet so far have been connected; the surface M: z2 - x2 - y2 = 1 (a hyper-boloid of two sheets) is not. Connectedness is a mild and natural conditionthat is sometimes included in the definition of surface.

The general definition of compactness is expressed in terms of open cov-erings. An open covering of a set A is a collection of open sets that covers Ain the sense that each point of A is in at least one of the sets.

A subset A in a space S (for us, either a Euclidean space or a surface) iscompact provided that given any open covering of A some finite number ofthe sets already covers A. In elementary calculus it is usually proved that aclosed interval I: a � t � b in R is compact, and this result extends to higherdimensions. In particular, any closed rectangle R: a � u � b, c � v � d in R2

is compact.We will need this abstract definition at a few crucial points, but in surface

theory the following concrete criterion is more useful.

7.2 Lemma A surface M is compact if and only if it can be covered bythe images of a finite number of 2-segments in M.

Proof. Suppose M is compact. For each point p in M, by using a coor-dinate patch containing p, we can construct a 2-segment whose image con-tains a neighborhood of p. The definition of compactness shows that afinite number of these neighborhoods already covers M; hence the corre-sponding 2-segments cover M.

The converse is an exercise in finiteness. First we show that the imagex(R) of a single 2-segment is compact. Recall that the definition of differ-entiability for 2-segments allows us to assume that x has been smoothlyextended over an open set containing the (closed) rectangle R.

Let {Ua} be an open covering of M. For each point r in R, one of thesesets, say Ur, contains x(r). Being differentiable, x is also continuous, so r


4.7 Topological Properties of Surfaces 185

has a neighborhood Nr that is carried into Ur by x. For all r, these neigh-borhoods Nr form an open covering of R. As mentioned above, the rec-tangle R is compact, so some finite number of these neighborhoods coverR. But this means that the corresponding original sets Ur (finite in number!)cover x(R); hence it is compact.

Now suppose that M is covered by the images of a finite number of 2-segments x1, . . . , xk. If {Ua} is an open covering of M, then we have justshown that a finite number of these sets suffice to cover the image of eachxi. Collecting these sets for i = 1, 2, . . . , k produces a finite number of setsUa that cover M. Thus M is compact. ◆

It follows at once from this proof that a region R in M is compact if it iscomposed of the images of finitely many 2-segments in M. For example,spheres are compact, since if the formula for the geographical patch(Example 2.2) is applied on the closed rectangle.

then this single 2-segment covers the entire sphere†. Similarly, the torus(Example 2.5) is compact, as is every surface of revolution whose profilecurve is closed.

The following lemma generalizes this fundamental fact: A continuous real-valued function on a closed rectangle R in the plane takes on a maximum atsome point of R.

7.3 Lemma A continuous function f on a compact region R in a surfaceM takes on a maximum at some point of M.

Proof. We show this in the only case we need: where R consists of theimages x1(R1), . . . , xk(Rk) of a finite number of 2-segments (for example,where R is an entire surface).

Since each xi is continuous, the composite functions f(xi): Ri Æ Rare all continuous. By the remark above, for each index i, there is a point(ui, vi) in Ri where f(xi) is a maximum. Let f(xj(uj, vj)) be the largest of thesefinite number k of maximum values; then evidently f(p) � f(xj(uj, vj)) forall p in M. ◆

R u v: - -p p p p� � � �, ,2 2

† Amazingly, every compact surface can be expressed as the image of a single 2-segment. See Ch. 1of [Ma].

This lemma is useful in proving noncompactness. For example, a cylinderC such as x2 + y2 = r2 is not compact since the coordinate function z isunbounded on C.

Finite size alone does not produce compactness. For example, the open diskD: x2 + y2 < 1 in the xy plane is itself a surface. Although D is bounded andhas finite area p, it is not compact since the function f = (1 - x2 - y2)-1 is continuous on D and does not have a maximum.

In general, a compact surface cannot have open edges, as D does, but mustbe smoothly closed up everywhere and finite in size—like a sphere or torus.

Roughly speaking, an orientable surface is one that is not twisted. Of themany equivalent definitions of orientability for surfaces, the following isperhaps the simplest.

7.4 Definition A surface M is orientable if there exists a differentiable (ormerely continuous) 2-form m on M that is nonzero at every point of M.

Recall that a 2-form is zero at a point p if it is zero on every pair of tangentvectors at p—or equivalently, on one linearly independent pair. Thus theplane R2 is orientable since du dv is a nonvanishing 2-form. This definition oforientability is somewhat mysterious, so for a surface M in R3 we give a moreintuitive description in terms of Euclidean geometry. A unit normal U on Mis a differentiable Euclidean vector field on M that has unit length and iseverywhere normal to M.

7.5 Proposition A surface M Ã R3 is orientable if and only if there existsa unit normal vector field on M. If M is connected as well as orientable, thereare exactly two unit normals, ±U.

Proof. We use the cross product of R3 to convert normal vector fieldsinto 2-forms, and vice versa.

Let U be a unit normal on M. If v and w are tangent vectors to M atp, define

Standard properties of the cross product show that m is a 2-form on M.When v and w are linearly independent, so are all three vectors, so m(v, w) π 0. Thus m is nonvanishing, which proves that M is orientable.

Conversely, suppose M is orientable, with m a nonvanishing 2-form.Again, nonvanishing implies that if v, w are linearly independent vectorsat a point p of M, then m(v, w) π 0. Now define

m v w p v w,( ) = ( ) ¥U • .



This formula is independent of the choice of v, w. Explicitly, for any othersuch pair v¢, w¢, it follows from Lemma 4.2 and the analogous formula forcross products that

Properties of the cross product show that Z(p) is nonzero and normal to M. The formula for cross product shows that Z is differentiable. ThusU = Z/||Z|| is the required unit normal.

If U is a unit normal on M, then so is -U. To show that there are noothers, let V also be a unit normal. At each point these (differentiable) unitvector fields are collinear, so the only values for the dot product V • U are+1 and -1. On a connected surface, a nonvanishing differentiable functioncannot change sign (Exercise 4), hence either V • U = +1 everywhere, so V = U, or V • U = -1 everywhere, so V = -U. ◆

For example, all spheres, cylinders, surfaces of revolution, and quadric sur-faces are orientable. It follows from Lemma 3.8 that every surface in R3 thatcan be defined implicitly is orientable.

However, nonorientable surfaces do exist in R3. The simplest example isthe famous Möbius band B, made from a strip of paper by giving it a halftwist, then gluing its ends together (Fig. 4.40). B is nonorientable since itcannot have a (differentiable) unit normal. To see this let g be a closed curve,as in Fig. 4.40, that runs once around the band with g (0) = g (1) = p. Nowsuppose a unit normal vector U at g (0) moves continuously around g. As thefigure shows, the twist in B forces the contradiction

U(p) = U(g (1)) = -U(g (0)) = -U(p).

¢ ¥ ¢¢ ¢( ) =

¥( )

v wv w

v wv wm m, ,

.

Z pv w

v w( ) =

¥( )m ,

.

FIG. 4.40

The last of the topological properties we consider will let us express for-mally the intuitive idea that a plane is simpler than a cylinder, and a sphereis simpler than a torus. The key is that in the cylinder or a torus there areclosed curves that cannot be continuously shrunk down to a point.

7.6 Definition A closed curve a in M is homotopic to a constant providedthere is a 2-segment x: R Æ M (called a homotopy) defined on

such that a is the base curve of x and the other three edge curves are con-stant at p = a(a) = a(b).

A curve such as a for which a(a) = a(b) holds but not necessarily a¢(a) =a¢(b) is often called a loop at p. Since the sides b and d of x are constant atp, for every 0 � v � 1 the u-parameter curve av(u) = x(u, v) is also a loop atp. As v varies from 0 to 1, the loop av varies continuously from a0 = a to thecurve a1 = g, which is constant at p.

It is easy to show that in the plane every loop is homotopic to a constant.Given a loop a: [a, b] Æ R2 at p in R2, use scalar multiplication in R2 to definex(u, v) = va(a) + (1 - v)a(u). Then

Hence x is a homotopy from a to a constant.

7.7 Definition A surface M is simply connected provided it is connectedand every loop in M is homotopic to a constant.

(This definition is valid for any manifold—and more generally.) The pre-ceding homotopy shows that the plane R2 is simply connected, and the sameformula works for any Euclidean space.

The 2-sphere S is also simply connected. Consider the following scheme ofproof. Let a be a loop in S at, say, the north pole of S. Pick a point q not ona. For simplicity, suppose q is the south pole. Now let x be the homotopyunder which each point of a moves due north along a great circle, reachingp in unit time. This x is a homotopy of a to a constant, as required.

But there is a difficulty here: finding the point q. In our usual case, wherea is differentiable, techniques from advanced calculus will show that there isalways a point q not on a. However, if a is merely continuous, it may actu-

x x p

x x p

u u u a u b

a v b v v

, and , for all , and

, , for all

0 1

0 1

( ) = ( ) ( ) =

( ) = ( ) =

a � �

� � .

R a u b v: � � � �, 0 1


ally fill the entire sphere. In this case, topological methods can be used todeform a slightly, making it no longer space-filling; then the scheme above isvalid.

To show that a given surface is simply connected, we can always try to con-struct the necessary homotopies; however, to show that a surface is not simplyconnected, indirect means are usually required. One of the most effectivederives from integration. Recall that a differential form f is closed if df = 0.

7.8 Lemma Let f be a closed 1-form on a surface M. If a loop a in Mis homotopic to a constant, then

Proof. Suppose x is a homotopy showing that a loop a is homotopic toa constant, say p. Now we apply Stokes’ theorem (Theorem 6.5). The inte-gral over a constant curve is zero, and df = 0, hence

◆

Now suppose we remove a single point, say the origin 0, from the (simplyconnected) plane R2. The loop a: [0, 2p] Æ C given by a(t) = (cos t, sin t)circles once around the missing point. It seems obvious that a cannot beshrunk down to a point in the punctured plane P = R2 - 0. The precedinglemma provides an easy way to prove it.

Exercise 4 of the preceding section shows that the 1-form

on P is closed and that its integral around a is 2p (these are easy computa-tions). By the lemma, a is not homotopic to a constant; hence P is not simplyconnected.

As noted earlier, since d 2 = 0, exact forms are always closed. However,closed forms need not be exact. For example, if the closed 1-form y wereexact, it would follow from Stokes’ theorem (6.5) that However, inan important special case, closed 1-forms are exact.

7.9 Lemma (Poincaré) On a simply connected surface, every closed 1-form is exact.

Proof. First we show that the integral of a closed 1-form is path indepen-

dent, that is, if a and b are curve segments from p to q, then

In fact, if -b is an orientation-reversing reparametrization of b, then a + (-b) = a - b is a loop. By simple connectedness, it is homotopic to aconstant. Thus, using Lemma 6.6,

f fa bÚ Ú= .

yaÚ = 0.

y =-+

x dy y dxx y2 2

0 = =ÚÚ Údf fax

.

faÚ = 0.



Now suppose f is a closed 1-form on a simply connected surfaceM. Pick a point p0 and define for any curve segment d from

p0 to p. Path independence makes f a well-defined function on M.To show that df = f, we must show that df(v) = f(v) for every tangent

vector v at a point p. This is equivalent to v[ f ] = f(v).Let a: [a, b] Æ M be a curve with initial velocity a ¢(a) = v. Then

d + a |[a, t] is a curve from p0 to a(t) (Fig. 4.41), so by the definition of f,

Taking the derivative with respect to t gives

When t = 0 this becomes v[ f ] = f(v), as required. ◆

Among the four properties we have discussed there are two direct impli-cations—both yielding orientability.

7.10 Theorem A compact surface in R3 is orientable.

This is an easy consequence of the following nontrivial topologicaltheorem, a 2-dimensional version of the Jordan Curve Theorem. If M is acompact surface in R3, then M separates R3 into two nonempty open sets: anexterior (the points that can escape to infinity) and an interior (the pointstrapped inside M). So we need only pick the unit normal vector at each p inM that points into, say, the exterior and apply Proposition 7.5. Thus orien-tation is at stake when in elementary calculus the “outward unit normal” isassigned to a surface in R3.

¢( )[ ] = ( )( )¢( ) = ¢( )( )a a f at f f t t .

f t f u dua t a

t

a f f ad a

( )( ) = = ( ) + ¢( )( )+ [ ]Ú Ú| ,

.p

f p( ) = Ú fd

0 = = + = -- -Ú Ú Ú Ú Úf f f f f

a b a b a b.

FIG. 4.41

7.11 Theorem A simply connected surface is orientable.

We defer the proof until Section 4 of Chapter 8. (Notice, for future refer-ence, that this theorem does not mention R3.)

A final note: Because the properties discussed in this section are topolog-ical, they can be defined solely in terms of open sets and continuous func-tions. However, differentiable versions are usually more practical for use ingeometry.

Exercises

1. Which of the following surfaces are compact and which are connected?(a) A sphere with one point removed.(b) The region z > 0 in M: z = xy.(c) A torus with the curve a(t) = x(t, t) removed. (See Ex. 2 of Sec. 4.3.)(d) The surface in Fig. 4.8.(e) M: x2 + y4 + z6 = 1.

2. Let F be a mapping of a surface M onto a surface N. Prove:(a) If M is connected, then N is connected.(b) If M is compact, then N is compact. (Try both the covering definitionand the criterion in Lem. 7.2.)

3. Let F: M Æ N be a regular mapping. Prove that if N is orientable, thenM is orientable.

4. Let f be a differentiable real-valued function on a connected surface.Prove:

(a) If df = 0, then f is constant.(b) If f is never zero then either f > 0 or f < 0.

5. Of the four basic types of surfaces of revolution (see Ex. 11 of Sec.5)—plane, sphere, cylinder, torus—which are,

(a) connected? (b) compact?(c) orientable? (d) simply connected?

A closed curve a in M is freely homotopic to a constant if the conditions onx in Definition 7.6 are weakened to b = d with only g required to be constant(Fig. 4.42). Then the v constant curves av are loops that move along the curveb = d as they shrink to p.

6. (a) If a loop a is freely homotopic to a constant via x, show that for any1-form f,



(b) If closed curves a and b in R2 - p are freely homotopic in R2 - p, showthat they have the same winding number about p.(c) A smooth disk D in a surface is the image of the unit disk x2 + y2 £ 1in R2 under a one-to-one regular map F. Show that the 2-segment

fills D and is a free homotopy of the (closed) boundary curve v Æ x(1, v) toa constant.

7. Let f be a closed 1-form and a a closed curve.

(a) Show that if either f is exact or a is freely homotopic to a constant.(b) Deduce that on a torus T the meridians and parallels are not freelyhomotopic to constants, and the closed 1-forms x and h of Exercise 9 ofSection 6 are not exact.

8. (Counterexamples.) Give examples to show that the following are false:(a) Converses of (a) and (b) of Exercise 2.(b) Exercise 3 with F not regular.(c) Converse of Exercise 3.

9. (a) If p is a point of a surface S, show that the set of all points of S thatcan be connected to p by a (piecewise smooth) curve in S is an open set ofS. (Hint: Each point of a surface has a neighborhood that is connected inthe sense of Def. 7.1.)

(b) Same as (a) but with can replaced by cannot.(c) For a surface M, show that Definition 7.1 (“path-connectedness”) isequivalent to the general topological definition of connectedness, namely:If U and M - U are open sets of M, and U contains at least one point,then M = U. (Use the corresponding property for a closed interval in R,a standard result of analysis.)

10. The Hausdorff axiom asserts that distinct points p π q have disjointneighborhoods. Prove:

faÚ = 0

x u v F u v u v u v, , , , ,( ) = ( ) ( )cos sin 0 1 0 2� � � � p

f faÚ ÚÚ= d

x.

FIG. 4.42

(a) R3 obeys the Hausdorff axiom. (The same proof works for all Rn.)(b) A surface M in R3 obeys the Hausdorff axiom.

11. If R is a compact region in a surface M, prove that R is a closed set ofM, that is, M - R is an open set. (Hint: To show that M - R is open, usethe preceding exercise and the fact that a finite intersection of neighborhoodsof p is again a neighborhood of p.)

12. Let M and N be surfaces in R3 such that M is contained in N.(a) If M is compact and N is connected, prove that M = N. (Hint: Showthat M is both closed and open in N.)(b) Give examples to show that (a) fails if either M is not compact or N isnot connected.(c) Deduce from (a) that if F: M Æ N is a local diffeomorphism with Mcompact and N connected, then F(M) = N.

4.8 Manifolds

Surfaces in R3 are a matter of everyday experience, so it is reasonable to inves-tigate them mathematically. But examining this concept with a critical eye,we may well ask whether there could not be surfaces in R4, or in Rn—or evensurfaces that are not contained in any Euclidean space. To devise a definitionfor such a surface, we must rely not on our direct experience of the real world,but on our mathematical experience of surfaces in R3. Thus we shall stripaway from Definition 1.2 every feature that involves R3. What is left will bejust a surface.

To begin with, a surface will be a set M: a collection of any objects what-soever. We call these objects the points of M, but as examples below will show,they definitely need not be the usual points of some Euclidean space. Anabstract patch in M will now be just a one-to-one function x: D Æ M froman open set D of R2 into the set M.

To get a workable definition of surface we must find a way to define whatit means for functions involving M to be differentiable. The key to thisproblem turns out to be the smooth overlap condition in Corollary 3.3. Toprove this condition is now a logical impossibility since R3 is gone, so in theusual fashion of mathematics, we make it an axiom.

8.1 Definition A surface is a set M furnished with a collection P ofabstract patches in M satisfying

(1) The covering axiom: The images of the patches in the collection P

cover M.

4.8 Manifolds 193


(2) The smooth overlap axiom: For any patches x, y in P, the compositefunctions y-1x and x-1y are Euclidean differentiable—and defined on opensets of R2.

This definition generalizes Definition 1.2: A surface in R3 is a surface inthis sense. However, there is a technical gap in this definition that requiresattention. First, for any patch x: D Æ M in a surface, define a set x(U ) to beopen provided U is open in D Ã R2. Then the open sets of M are all unionsof such sets. (This is consistent with the case M Ã R3, since there x and x-1

are continuous.)Examples like that in Exercise 11 show that for the open sets to behave

properly we must add another axiom to the definition of surface.

(3) The Hausdorff axiom: For any points p π q in M there exist disjoint(that is, nonoverlapping) patches x and y with p in x(D) and q in y(E).

Here is an example of an important surface that, as we will soon see,cannot be found in R3.

8.2 Example The projective plane P. Starting from the unit sphere S inR3 we construct P by identifying antipodal points in S, that is, by consideringp and -p to be the same point of P (Fig. 4.43). Formally, this means that theset P consists of all antipodal pairs {p, -p} of points in S. Order is not relevant here; that is, {p, -p} = {-p, p}. (Working with the projective planeoften involves looking back and forth between antipodal points.)

There are two important mappings associated with P: the antipodal map A(p) = -p on S and the projection F(p) = {p, -p} of S onto P. Note thatFA = F.

Call a patch x in S “small” if it is contained in a single open hemisphere.Then the composite function Fx is one-to-one, and is thus an abstract patch.The collection of all such abstract patches makes P a surface. In fact, the cov-ering condition (1) is clear, and the Hausdorff axiom derives from the cor-responding property for Euclidean spaces. The smooth overlap axiom (2) canbe checked as follows.

FIG. 4.43

Suppose Fx and Fy overlap in S; that is, their images have points incommon. If x and y overlap in S, then (Fy)-1Fx = y-1x, which by Corollary3.3 is differentiable and defined on an open set. On the other hand, if x andy do not overlap, replace y by Ay. Then x and Ay do overlap, so the previ-ous argument applies.

To emphasize the distinction between a surface in R3 and the generalnotion of surface defined above, we sometimes call the latter an abstractsurface.

To get as many patches as possible in an abstract surface M we alwaysunderstand that its patch collection P has been enlarged to include all theabstract patches in M that overlap smoothly with those originally in P. Weemphasize that abstract surfaces M1 and M2 with the same set of points arenevertheless different surfaces if their (enlarged) patch collections P1 and P2

are different.There is essentially only one problem to solve in establishing calculus on

an abstract surface M, and that is to define the velocity of a curve in M. Inthe old definition a velocity vector was a tangent vector to R3, so somethingnew is needed. For everything else—differentiable functions, curves them-selves, tangent vectors, vector fields, differentiable forms, and so on—the def-initions and theorems given for surfaces in R3 apply without change. It isnecessary to tinker with a few proofs, but no serious problems arise.

It makes little difference what we define velocity to be—provided the new definition produces the same essential behavior as before. The most efficient choice is based on the directional derivative property in Lemma 4.6,Chapter 1.

8.3 Definition Let a: I Æ M be a curve in an abstract surface M. Foreach t in I the velocity vector a ¢(t) is the function such that

for every differentiable real-valued function f on M.

Thus a ¢(t) is a real-valued function whose domain is the set F of all real-valued functions on M. This is all we need to generalize the calculus on sur-faces in R3 to the case of an abstract surface.

We now have a calculus for Rn (Chapter 1) and another one for surfaces.These are strictly analogous, but analogies in mathematics, though useful initially, can be annoying in the long run. What we need is a single calculusof which these two will be special cases. The most general object on which

¢( )[ ] =( ) ( )a

at f

d fdt

t

4.8 Manifolds 195

calculus can be conducted is called a manifold. It is simply an abstract surfaceof arbitrary dimension n.

8.4 Definition An n-dimensional manifold M is a set furnished with a col-lection P of abstract patches (one-to-one functions x: D Æ M, D an open setin Rn) satisfying

(1) The covering property: The images of the patches in the collection P

cover M.(2) The smooth overlap property: For any patches x, y in P, the compos-

ite functions y-1x and x-1y are Euclidean differentiable—and defined on opensets of Rn.

(3) The Hausdorff property: For any points p π q in M there are disjointpatches x and y with p in x(D) and q in y(E).

Thus a surface (Definition 8.1) is just a 2-dimensional manifold. As before,Euclidean n-space Rn is an n-dimensional manifold whose (initial) patch col-lection consists only of the identity map.

To keep this definition as close as possible to that of a surface in R3, wehave deviated somewhat from the standard definition of manifold in whichit is the inverse functions x-1: x(D) Æ D that are axiomatized.

The calculus of an arbitrary n-dimensional manifold is defined in the sameway as for n = 2. Usually we need only replace i = 1, 2 by i = 1, 2, . . . , n.Differential forms on an n-dimensional manifold have the same general prop-erties as in the case n = 2, which we have explored in Sections 4, 5, and 6.But there are p-forms for 0 � p � n, so when n is large, the algebra becomesmore complicated.

Wherever calculus appears in mathematics and its applications, manifoldswill be found, and higher dimensional manifolds turn out to be important inproblems—both pure and applied—that initially seem to involve only dimen-sions 2 or 3. For example, here is a 4-dimensional manifold that has alreadyappeared, implicitly at least, in this chapter.

8.5 Example The tangent bundle of a surface. For a surface M, let T(M)be the set of all tangent vectors to M at all points of M. (For simplicity weassume M is a surface in R3, but it could just as well be an abstract surfaceor, indeed, a manifold of any dimension.) Since M has dimension 2 and eachtangent space Tp(M) has dimension 2, we anticipate that T(M) will havedimension 4.

To get a natural patch collection for T(M) we derive from each patch x inM an abstract patch in T(M). Given x: D Æ M, let be the open set in R4Dx


4.8 Manifolds 197

consisting of all points (p1, p2, p3, p4) for which (p1, p2) is in D. Then let bethe function Æ T(M) given by

(In Fig. 4.44 we identify R2 with the x1x2 plane of R4 and deal as best we canwith dimension 4.)

Using Exercise 3 of Section 3 and the proof of Lemma 3.6, it is not diffi-cult to check that each such function is one-to-one and that the collectionof all such patches satisfies the conditions in Definition 8.4. Thus T(M) is a4-dimensional manifold, called the tangent bundle of M.

Exercises

1. Prove that a surface M is nonorientable if there is a smoothly closedcurve a: [a, b] Æ M and a tangent vector field Y on a such that

(i) Y and a ¢ are linearly independent at every point, and(ii) Y(b) = -Y(a).

(Hint: Assume M is orientable and deduce a contradiction.)

2. Establish the following properties of the projective plane P.(a) If F: S Æ P is the projection, then each tangent vector to P is the imageunder F* of exactly two tangent vectors to S, these of the form vp and -v-p.(b) P is compact, connected, and nonorientable—hence P is not diffeo-morphic to any surface in R3.

3. (a) Prove that the tangent bundle T(M) of a surface is a manifold. (If xand y are overlapping patches in M, find an explicit formula for -1 .)

(b) If M is the image of a single patch x: R2 Æ M, show that the tangentbundle of M is diffeomorphic to R4.

xy

x

˜ .x x xp p p p p p p p p pu v1 2 3 4 3 1 2 4 1 2, , , , ,( ) = ( ) + ( )D

x

FIG. 4.44

4. A surface M in R3 is closed if it is a closed set of R3, that is, R3 - M isan open set. (Confusingly, closed surface has sometimes been defined byanalogy with closed curve to mean compact surface.)

(a) Prove that every surface given in the implicit form M: g = c is closed.(b) Prove that a compact surface in R3 is closed. (Hint: Use the scheme ofEx. 11 of Sec. 7.)(c) Give an example of a closed surface in R3 that is not compact.

5. A surface M in R3 is bounded provided there is a number R > 0 such that||p|| � R for all p Œ M.

(a) Prove that a compact surface M Ã R3 is bounded.(b) Give an example of a surface in R3 that is bounded but not compact.It follows from this exercise and Exercise 4 that if M Ã R3 is compact, itis closed and bounded. The converse is true but is more difficult to prove(see [Mu]).

6. Let M be the set of all the unit normal vectors on a surface M in R3. (Sothere are two points of M for each point of M.) For each patch x in M wedefine two patches in M, namely,

(a) Show that the set of all such patches makes M a surface, called the ori-entation covering surface of M.(b) Describe the orientation covering surface of the unit sphere S Ã R3 andof the torus T Ã R3.(c) If a connected surface M in R3 is orientable, show that M consists oftwo diffeomorphic copies of M. (Hint: M has smooth unit normals ±U.)(d) The natural map ±U Æ p of M onto M is regular.

7. (Continuation.) If a connected surface M in R3 is nonorientable, showthat M is (i) connected and (ii) orientable. (Thus nonorientability can becured by doubling. For an example, see Ex. 10.)(Hints: (i) If a: [a, b] Æ M is a curve, then any unit normal vector at a(a)can be moved differentiably along a as a unit normal Ua(t)—thus giving acurve in M. Assume this fact: if M is nonorientable, there exists a loop a inM with Ua(b) = -Ua(a).

(ii) Any patch z determines a unit vector field Uz = zu ¥ zv/||zu ¥ zv||. Ifpatches x and y in M meet, but Uy = -Ux on the overlap, then x and y donot meet.)

8. A Möbius band B can be constructed as a ruled surface by

x u v u v u v, , with, say, ,( ) = ( ) + ( ) - < <b d 1 3 1 3

xx xx x±( ) = ±

¥¥

u v u v

u v

, .


where b(u) = (cosu, sinu, 0) and

(The ruling makes only a half turn as it traverses the circle b.)Show that x is one-to-one and regular, with unit normal

9. (Continuation, computer graphics.)(a) Plot the Möbius band B.(b) Plot a surface that represents B with its central circle b removed (for clarity, remove a small band around b). Is this surface connected? orientable?

Although a point of M is a vector in R3, not a point of R3, in favorablecases M can be turned into a surface in R3 by mapping each Up in M tothe point p + eUp (Euclidean coordinates) in R3 for some small e > 0. (Fore = 1, the point would be the tip of the “arrow” Up.)

10. (Continuation.)(a) Using the scheme above, plot the orientation covering surface B of theMöbius band B. (Take e = 1/4.)(b) By inspection, is B connected? orientable? How is B related to thesurface in (b) of the preceding exercise?

11. (Plane with two origins.) Let Z consist of all ordered pairs of realnumbers and one additional point 0*. Let x and y be the functions from R2

to Z such that

but

(a) Show that the abstract patches x and y constitute a patch collectionthat satisfies the first two conditions in Definition 8.1, but not the Haus-dorff axiom. Without the Hausdorff axiom, strange things can happen.For example, prove:(b) A convergent sequence in Z can have two limits.(c) The function F: Z Æ Z that reverses 0 and 0*, leaving all other pointsfixed, is a differentiable mapping.

x 0 y 00 0 0 0 0 0, , and , *( ) = = ( ) ( ) = .

x yu v u v u v u v, , , if , , ,( ) = ( ) = ( ) ( ) π ( )0 0

U u vu v u

u vu

u

,,

,( ) =

( ) ¥ ( )( )

xx

d.

d uu

uu

uu( ) = Ê

Ëˆ¯cos cos cos sin sin .

2 2 2, ,

4.8 Manifolds 199

12. (a) Given a one-to-one function H from a manifold M onto an arbi-trary set A, prove there is a unique way to make A a manifold so that Hbecomes a diffeomorphism. (Hint: Diffeomorphisms move patches topatches.)

(b) In each of the following cases, find natural choices of H and M thatmake the set a manifold.(i) The set of all 2 ¥ 2 real symmetric matrices.(ii) The set of all circles in R2.(iii) The set of all great circles on a sphere S.(iv) The set of all (finite) closed intervals in R.

13. (Integral curves.) A curve a in M is an integral curve of a vector field Von M provided a ¢(t) = V(a(t)) for all t. Thus an integral curve has at eachpoint the velocity prescribed by V. If a(0) = p, we say that a starts at p.

(a) In R2, show that the curve a(t) = (u(t), v(t)) is an integral curve ofV = f1U1 + f2U2 starting at (a, b) Œ R2 if and only if

The theory of differential equations guarantees that there is a unique solu-tion for a.

(b) Find an explicit formula for the integral curve of V = -u2U1 + uvU2

on R2 that starts at the point (1, -1). (The differential equations involvedcan be solved by elementary methods since one of them is particularlysimple.)(c) Sketch (by hand or by computer) the integral curve a on suitable inter-vals A < t < -1 and -1 < t < B.

14. (Continuation.) Show that every vector field V on a surface M has anintegral curve b starting at any given point p. Specifically, if x: D Æ M is apatch with x(a, b) = p, and is the vector field on D such that x*( ) = V,show that b(t) = x(u(t), v(t)), where (u(t), v(t)) is the integral curve of start-ing at (a, b).

15. (Cartesian products.) For any sets A and B the Cartesian product A¥Bconsists of all ordered pairs (a, b), with a in A and b in B. If x: D Æ M and y: E Æ N are patches in surfaces M and N, define x¥y: D¥E Æ M¥N by

Show that x¥y is an abstract patch and that the collection P of all suchpatches makes M¥N a 4-dimensional manifold. M¥N is called the Cartesianproduct of M and N.

x y x y¥( ) ¢ ¢( ) = ( ) ¢ ¢( )( )u v u v u v u v, , , , , , .

VVV

¢ = ( ) ( ) =¢ = ( ) ( ) =

u f u v u a

v f u v v b1

2

0

0

, , ,

, , .


The same scheme works for any two manifolds. It derives from the way thex axis and y axis produce the xy plane; indeed, R¥R is precisely R2.

16. If M is an abstract surface, a proper imbedding of M into R3 is a one-to-one regular mapping F: M Æ R3 such that the inverse function F-1:F(M) Æ M is continuous. Prove that the image F(M) of a proper imbeddingis a surface in R3 (Def. 1.2) and that it is diffeomorphic to M.

If F: M Æ R3 is merely regular, then F is an immersion of M into R3, andthe image F(M) is often called an “immersed surface,” even though it can cutacross itself and hence not satisfy Definition 1.2.

4.9 Summary

The discovery of calculus made it possible to study arbitrary curved surfacesM in R3. Initially this was done mostly in terms of the natural Euclideancoordinates {x, y, z} of R3. However, it gradually became clear that in manycontexts, coordinates {u, v} in the surface itself were more efficient. Thus atwo-dimensional calculus was developed for surfaces, one that remains valideven if the surface is not contained in R3.

Along with the Euclidean spaces, such surfaces are prime examples of thegeneral notion of manifold. The calculus of any manifold involves differen-tiable functions, vector fields, differential forms, mappings—and variousoperations of differentiation and integration. These features are all preservedin a suitable sense by diffeomorphisms—indeed, this criterion gives a formaldefinition of manifold theory.

4.9 Summary 201

▼

▲Chapter 5

Shape Operators

202

In Chapter 2 we measured the shape of a curve in R3 by its curvature andtorsion functions. Now we consider the analogous measurement problem forsurfaces. It turns out that the shape of a surface M in R3 is described infini-tesimally by a certain linear operator S defined on each of the tangent planesof M. As with curves, to say that two surfaces in R3 have the same shape meanssimply that they are congruent. And just as with curves, we shall justify ourinfinitesimal measurements by proving that two surfaces with “the same”shape operators are, in fact, congruent. The algebraic invariants (determinant,trace, . . .) of its shape operators thus have geometric meaning for the surfaceM. We investigate this matter in detail and find efficient ways to compute theseinvariants, which we test on a number of geometrically interesting surfaces.

From now on, the notation M Ã R3 means a connected surface M in R3 asdefined in Chapter 4.

5.1 The Shape Operator of M Ã R3

Suppose that Z is a Euclidean vector field (Definition 3.7 of Chapter 4) on asurface M in R3. Although Z is defined only at points of M, the covariantderivative �vZ (Chapter 2, Section 5) still makes sense as long as v is tangentto M. As usual, �vZ is the rate of change of Z in the v direction, and thereare two main ways to compute it.

Method 1. Let a be a curve in M that has initial velocity a ¢(0) = v. LetZa be the restriction of Z to a, that is, the vector field t Æ Z(a(t)) on a. Then

where the derivative is that of Chapter 2, Section 2.

— = ( )¢ ( )vZ Za 0 ,

5.1 The Shape Operator of M Ã R3 203

Method 2. Express Z in terms of the natural frame field of R3 by

Then

where the directional derivative is that of Definition 3.10 in Chapter 4.It is easy to show that these two methods give the same result. In fact, since

,

(Compare Lemma 5.2 of Chapter 2.) Note that even if Z is a tangent vectorfield, the covariant derivative �vZ need not be tangent to M.

If M is an orientable surface, Proposition 7.5 of Chapter 4 shows that thereis always a (differentiable) unit normal vector field U on the entire surface,and in fact—since M is now assumed connected—there are exactly two, ±U.Even if M is not orientable, unit normals ±U are available locally, since asmall region around any point is orientable. In fact, we will find explicit for-mulas for U on the image x(D) Ã M of any patch.

We are now in a position to find a mathematical measurement of the shapeof a surface in R3.

1.1 Definition If p is a point of M, then for each tangent vector v to Mat p, let

where U is a unit normal vector field on a neighborhood of p in M. Sp iscalled the shape operator of M at p derived from U.† (Fig. 5.1.)

S Up vv( ) = -— ,

Z z U z Ui i i ia a( )¢ ( ) = ( )¢ ( ) = [ ]Â Â0 0 v .

Z zUi i= Â

— = [ ]Âv i iZ z Uv ,

Z zUi i= Â .

Sp(v) v

FIG. 5.1

† The minus sign artificially introduced in this definition will sharply reduce the total number ofminus signs needed later on.

The tangent plane of M at any point q consists of all Euclidean vectorsorthogonal to U(q). Thus the rate of change �vU of U in the v direction tellshow the tangent planes of M are varying in the v direction—and this givesan infinitesimal description of the way M itself is curving in R3.

Note that if U is replaced by -U, then Sp changes to -Sp.

1.2 Lemma For each point p of M Ã R3, the shape operator is a linearoperator

on the tangent plane of M at p.

Proof. In Definition 1.1, U is a unit vector field, so U • U = 1. Thus bya Leibnizian property of covariant derivatives,

where v is tangent to M at p. Since U is also a normal vector field, it followsthat Sp(v) is tangent to M at p. Thus Sp is a function from Tp(M) to Tp(M).(It is to emphasize this that we use the term “operator” instead of“transformation.”)

The linearity of Sp is a consequence of a linearity property of covari-ant derivatives:

◆

At each point p of M Ã R3 there are actually two shape operators, ±Sp,derived from the two unit normals ±U near p. We shall refer to all of these,collectively, as the shape operator S of M. Thus if a choice of unit normal isnot specified, there is a relatively harmless ambiguity of sign.

1.3 Example Shape operators of some surfaces in R3.(1) Let S be the sphere of radius r consisting of all points p of R3 with

|| p || = r. Let U be the outward normal on S. Now as U moves away fromany point p in the direction v, evidently U topples forward in the exact direc-tion of v itself (Fig. 5.2). Thus S(v) must have the form -cv.

In fact, using gradients as in Example 3.9 of Chapter 4, we find

Ur

xUi i= Â1.

S a b U a U b U

aS bS

p av bw v w

p p

v w

v w

+( ) = -— = - — + —( )= ( ) + ( )

+

.

0 2 2= [ ] = — ( ) = - ( ) ( )v p v pU U U U S Uv p• • • ,

S T M T Mp p p: ( ) Æ ( )

204 5. Shape Operators


But then

Thus for all v. So the shape operator S is merely scalarmultiplication by . This uniformity in S reflects the roundness ofspheres: They bend the same way in all directions at all points.

(2) Let P be a plane in R3. A unit normal vector field U on P is evidentlyparallel in R3 (constant Euclidean coordinates) (Fig. 5.3). Hence

for all tangent vectors v to P. Thus the shape operator is identically zero,which is to be expected, since planes do not bend at all.

(3) Let C be the circular cylinder x2 + y2 = r2 in R3. At any point p of C,let e1 and e2 be unit tangent vectors, with e1 tangent to the ruling of the cylin-der through p, and e2 tangent to the cross-sectional circle. Use the outwardnormal U as indicated in Fig. 5.4.

Now, when U moves from p in the e1 direction, it stays parallel to itself justas on a plane; hence S(e1) = 0. When U moves in the e2 direction, it topples

S Uvv( ) = -— = 0

-1/rS rv v( ) = - /

— = [ ] ( ) = -Âv i iUr

x Ur

1v p

v.

FIG. 5.2

FIG. 5.3

FIG. 5.4


forward exactly as on a sphere of radius r; hence S(e2) = -e2/r. In this way Sdescribes the “half-flat, half-round” shape of a cylinder.

(4) The saddle surface M: z = xy. For the moment we investigate S only atp = (0, 0, 0) in M. Since the x and y axes of R3 lie in M, the vectors

u1 = (1, 0, 0) and u2 = (0, 1, 0)

are tangent to M at p. We use the “upward” unit normal U, which at p is (0, 0, 1). Along the x axis, U stays orthogonal to the x axis, and as it pro-ceeds in the u1 direction, U swings from left to right (Fig. 5.5).

FIG. 5.5

In fact, a routine computation (Exercise 3) shows that �u1U = -u2. Similarly,

we find �u2U = -u1. Thus the shape operator of M at p is given by the formula

These examples clarify the analogy between the shape operator of a surfaceand the curvature and torsion of a curve. In the case of a curve, there is onlyone direction to move, and k and t measure the rate of change of the unitvector fields T and B (hence N). For a surface, only one unit vector field isintrinsically determined—the unit normal U. Furthermore, at each point,there is now a whole plane of directions in which U can move, so that ratesof change of U are measured, not numerically, but by the linear operators S.

1.4 Lemma For each point p of M Ã R3, the shape operator

is a symmetric linear operator; that is,

for any pair of tangent vectors to M at p.

S Sv w w v( ) = ( )• •

S T M T Mp p: ( ) Æ ( )

S a b b au u u u1 2 1 2+( ) = + .

We postpone the proof of this crucial fact to Section 4, where it occursnaturally in the course of general computations.

From the viewpoint of linear algebra, a symmetric linear operator on atwo-dimensional vector space is a very simple object indeed. For a shapeoperator, its eigenvalues and eigenvectors, its trace and determinant, all turnout to have geometric meaning of first importance for the surface M Ã R3.

Exercises

1. Let a be a curve in M Ã R3. If U is a unit normal of M restricted to thecurve a, show that S(a ¢) = -U ¢.

2. Consider the surface M: z = f(x, y), where

(The subscripts indicate partial derivatives.) Show that(a) The vectors u1 = U1(0) and u2 = U2(0) are tangent to M at the origin 0,and

is a unit normal vector field on M.(b) S(u1) = fxx(0, 0)u1 + fxy(0, 0)u2,

S(u2) = fyx(0, 0)u1 + fyy(0, 0)u2.

(Note: The square root in the denominator is no real problem here becauseof the special character of f at (0, 0). In general, direct computation of S isawkward, and in Section 4 we shall establish indirect ways of getting at it.)

3. (Continuation.) In each case, express S(au1 + bu2) in terms of u1 and u2,and determine the rank of S at 0 (rank S is dimension of image S: 0, 1, or2).

(a) z = xy. (b) z = 2x2 + y2.(c) z = (x + y)2. (d) z = xy2.

4. Let M be a surface in R3 oriented by a unit normal vector field

Then the Gauss map G: M Æ S of M sends each point p of M to the point(g1(p), g2(p), g3(p)) of the unit sphere S. Pictorially: Move U(p) to the originby parallel motion; there it points to G(p) (Fig. 5.6).

Thus G completely describes the turning of U as it traverses M.

U g U g U g U= + +1 1 2 2 3 3.

Uf U f U U

f f

x y

x y

=- - +

+ +1 2 3

2 21

f f fx y0 0 0 0 0 0 0, , ,( ) = ( ) = ( ) = .



For each of the following surfaces, describe the image G(M ) of the Gaussmap in the sphere S (use either normal):

(a) Cylinder, x2 + y2 = r2.(b) Cone,(c) Plane, x + y + z = 0.(d) Sphere, (x - 1)2 + y2 + (z + 2)2 = 1.

5. Let G: T Æ S be the Gauss map of the torus T (Ex. 2.5 of Ch. 4) derivedfrom its outward unit normal U. What are the image curves under G of themeridians and parallels of T? Which points of S are the image of exactly twopoints of T?

6. Let G: M Æ S be the Gauss map of the saddle surface M: z = xy derivedfrom the unit normal U obtained as in Exercise 2. What is the image underG of one of the straight lines, y = constant, in M? How much of the sphereis covered by the entire image G(M )?

7. Show that the shape operator of M is (minus) the tangent map of itsGauss map: If S and G: M Æ S are both derived from U, then S(v) and -G*(v) are parallel for every tangent vector v to M.

8. An orientable surface has two Gauss maps derived from its two unitnormals. Show that they differ only by the antipodal mapping of S (Ex. 8.2of Ch. 4). Define a Gauss-type mapping for a nonorientable surface in R3.

9. If V is a tangent vector field on M (with unit normal U), then S(V) isthe tangent vector field on M whose value at each point p is Sp(V(p)). Showthat if W is also tangent to M, then

Deduce that the symmetry of S is equivalent to the assertion that the bracket

of two tangent vector fields is again a tangent vector field.

V W W VV W,[ ] = — - —

S V W W UV( ) = —• • .

z x y= +2 2 .

FIG. 5.6

5.2 Normal Curvature 209

5.2 Normal Curvature

Throughout this section we shall work in a region of M Ã R3 that has beenoriented by the choice of a unit normal vector field U, and we use the shapeoperator S derived from U.

The shape of a surface in R3 influences the shape of the curves in M.

2.1 Lemma If a is a curve in M Ã R3, then

Proof. Since a is in M, its velocity a ¢ is always tangent to M. Thus

a ¢ • U = 0,

where U is restricted to the curve a. Differentiation yields

But from Section 1, we know that S(a ¢) = -U¢. Hence

◆

Geometric interpretation: at each point, a ≤ • U is the component of accel-eration a ≤ normal to the surface M (Fig. 5.7). The lemma shows that thiscomponent depends only on the velocity a¢ and the shape operator of M.Thus all curves in M with a given velocity v at point p will have the same normalcomponent of acceleration at p, namely, S(v) • v. This is the component ofacceleration that the bending of M in R3 forces them to have.

Thus if v is standardized by reducing it to a unit vector u, we get a mea-surement of the way M is bent in the u direction.

2.2 Definition Let u be a unit vector tangent to M Ã R3 at a point p.Then the number k(u) = S(u) • u is called the normal curvature of M in the udirection.

¢¢ = - ¢ ¢ = ¢( ) ¢a a a a• • •U U S

¢¢ + ¢ ¢ =a a• • .U U 0

¢¢ = ¢( ) ¢a a a• • .U S

FIG. 5.7


In this context, we define a tangent direction to M at p to be a one-dimensional subspace L of Tp(M), that is, a line through the zero vector(located for intuitive purposes at p). Any nonzero tangent vector at p deter-mines a direction L, but we prefer to use one of the two unit vectors ±uin L. Note that

Thus, although we evaluate k on unit vectors, it is, in effect, a real-valuedfunction on the set of all tangent directions to M.

Given a unit tangent vector u to M at p, let a be a unit-speed curve in Mwith initial velocity a ¢(0) = u. Using the Frenet apparatus of a, the preced-ing lemma gives

Thus the normal curvature of M in the u direction is k (0)cosJ, where k(0)is the curvature of a at a(0) = p, and J is the angle between the principalnormal N(0) and the surface normal U(p), as in Fig. 5.8.

Given u, there is a natural way to choose the curve so that J is 0 or p. Infact, if P is the plane determined by u and U(p), then P cuts from M (nearp) a curve s called the normal section of M in the u direction. If we give sunit-speed parametrization with s ¢(0) = u, then N(0) = ±U(p), since

s ≤(0) = k (0)N(0)

is orthogonal to s ¢(0) = u and tangent to P. So for a normal section in the udirection (Fig. 5.9),

k N Uu p( ) = ( ) ( ) ( ) = ± ( )k ks s0 0 0• .

k S U N Uu u u p p( ) = ( ) = ¢¢( ) ( ) = ( ) ( ) ( )= ( )

• • •

cos .

a k

k J

0 0 0

0

k S S ku u u u u u( ) = ( ) = -( ) -( ) = -( )• • .

FIG. 5.8


Thus it is possible to make a reasonable estimate of the normal curvaturesin various directions on a surface M Ã R3 by picturing what the corre-sponding normal sections would look like. We know that the principalnormal N of a curve tells in which direction it is turning. Thus the preced-ing discussion gives geometric meaning to the sign of the normal curvaturek(u) (relative to our fixed choice of U).

(1) If k(u) > 0, then N(0) = U(p), so the normal section s is bendingtoward U(p) at p (Fig. 5.10). Thus in the u direction the surface M is bendingtoward U(p).

(2) If k(u) < 0, then N(0) = -U(p), so the normal section s is bendingaway from U(p) at p. Thus in the u direction M is bending away from U(p)(Fig. 5.11).

(3) If k(u) = 0, then ks(0) = 0 and N(0) is undefined. Here the normalsection s is not turning at s(0) = p. We cannot conclude that in the u direc-tion M is not bending at all, since k might be zero only at s(0) = p. But wecan conclude that its rate of bending is unusually small.

In different directions at a fixed point p, the surface may bend in quite dif-ferent ways. For example, consider the saddle surface z = xy in Example 1.3.If we identify the tangent plane of M at p = (0, 0, 0) with the xy plane of

FIG. 5.9

FIG. 5.10 FIG. 5.11


R3, then clearly the normal curvature in the direction of the x and y axes iszero, since the normal sections are straight lines. However, Fig. 5.5 shows thatin the tangent direction given by the line y = x, the normal curvature is pos-itive, for the normal section is a parabola bending upward. (U(p) = (0, 0, 1)is “upward.”) But in the direction of the line y = -x, normal curvature isnegative, since this parabola bends downward.

Let us now fix a point p of M Ã R3 and imagine that a unit tangent vectoru at p revolves, sweeping out the unit circle in the tangent plane Tp(M). Fromthe corresponding normal sections, we get a moving picture of the way M isbending in every direction at p (Fig. 5.12).

2.3 Definition Let p be a point of M Ã R3. The maximum and minimumvalues of the normal curvature k(u) of M at p are called the principal curva-tures of M at p, and are denoted by k1 and k2. The directions in which theseextreme values occur are called principal directions of M at p. Unit vectorsin these directions are called principal vectors of M at p.

Using the normal-section scheme discussed above, it is often fairly easy topick out the directions of maximum and minimum bending. For example, ifwe use the outward normal (U) on a circular cylinder C as in Fig. 5.4, thenthe normal sections of C all bend away from U, so k(u) � 0. Furthermore,it is reasonably clear that the maximum value k1 = 0 occurs only in the direc-tion e1 of a ruling; minimum value k2 < 0 occurs only in the direction e2

tangent to a cross-section.An interesting special case occurs at points p for which k1 = k2. The

maximum and minimum normal curvature being equal, it follows that k(u)is constant: M bends the same amount in all directions at p (so all directionsare principal).

2.4 Definition A point p of M Ã R3 is umbilic provided the normal cur-vature k(u) is constant on all unit tangent vectors u at p.

FIG. 5.12


For example, what we found in (1) of Example 1.3 was that every point ofthe sphere S is umbilic, with

2.5 Theorem (1) If p is an umbilic point of M Ã R3, then the shape oper-ator S at p is just scalar multiplication by k = k1 = k2.

(2) If p is a nonumbilic point, k1 π k2, then there are exactly two princi-pal directions, and these are orthogonal. Furthermore, if e1 and e2 are prin-cipal vectors in these directions, then

In short, the principal curvatures of M at p are the eigenvalues of S, andthe principal vectors of M at p are the eigenvectors of S.

Proof. Suppose that k takes on its maximum value k1 at e1, so

Let e2 be merely a unit tangent vector orthogonal to e1 (presently we shallshow that it is also a principal vector).

If u is any unit tangent vector at p, we write

where c = cosJ, s = sinJ (Fig. 5.13). Thus normal curvature k at p becomesa function on the real line: k(J) = k(u(J)).

For 1 � i, j � 2, let Sij be the number S(ei) · ej. Note that S11 = k1, andby the symmetry of the shape operator, S12 = S21. We compute

(1)

Hence

k S c s c s

c S scS s S

J( ) = +( ) +( )= + +

e e e e1 2 1 2

211 12

2222

•

.

u u e e= ( ) = +J c s1 2 ,

k k S1 1 1 1= ( ) = ( )e e e• .

S k S ke e e e1 1 1 2 2 2( ) = ( ) =, .

k k r1 2 1= = - .

FIG. 5.13


(2)

If J = 0, then c = 1 and s = 0, so u(0) = e1. Thus, by assumption, k(J)is a maximum at J = 0, so (dk/dJ)(0) = 0. It follows immediately from (2)that S12 = 0.

Since e1, e2 is an orthonormal basis for Tp(M), we deduce by orthonor-mal expansion that

(3)

Now if p is umbilic, then S22 = k(e2) is the same as S11 = k(e1) = k1, so (3)shows that S is scalar multiplication by k1 = k2.

If p is not umbilic, we look back at (1), which has become

(4)

Since k1 is the maximum value of k(J), and k(J) is now nonconstant, itfollows that k1 > S22. But then (4) shows: (a) the maximum value k1 is takenon only when c = ±1, s = 0, that is, in the e1 direction; and (b) the minimumvalue k2 is S22, and is taken on only when c = 0, s = ±1 that is, in the e2 direc-tion. This proves the second assertion in the theorem, since (3) now reads:

◆

Contained in the preceding proof is Euler’s formula for the normal cur-vature of M in all directions at p.

2.6 Corollary Let k1, k2 and e1, e2 be the principal curvatures and vectorsof M Ã R3 at p. Then if u = cosJ e1 + sinJ e2, the normal curvature of Min the u direction is (Fig. 5.13)

Here is another way to show how the principal curvatures k1 and k2 controlthe shape of M near an arbitrary point p. Since the position of M in R3 is ofno importance, we can assume that (1) p is at the origin of R3, (2) the tangentplane Tp(M) is the xy plane of R3, and (3) the x and y axes are the principaldirections. Near p, M can be expressed as M: z = f(x, y), as shown in Fig.5.14, and the idea is to construct an approximation of M near p by using onlyterms up to quadratic in the Taylor expansion of the function f. Now (1) and(2) imply f 0 = f 0

x = f 0y = 0, where the subscripts indicate partial derivatives

and the superscript zero denotes evaluation at x = 0, y = 0. Thus the qua-dratic approximation of f near (0, 0) reduces to

k k ku( ) = +12

22cos sin .J J

S k ke e e e1 1 1 2 2 2( ) = ( ) =, S .

k c k s SJ( ) = +21

222.

S S Se e e e1 11 1 2 22 2( ) = ( ) =, S .

dkd

sc S S c S SJ

J( ) = -( ) + -( )2 222 112 2

12.


In Exercise 2 of Section 1 we found that for the tangent vectors

at p = 0,

By condition (3) above, u1 and u2 are principal vectors, so it follows fromTheorem 2.5 that k1 = f 0

xx, k2 = f 0yy, and f 0

xy = 0.Substituting these values in the quadratic approximation of f, we conclude

that the shape of M near p is approximately the same as that of the surface

near 0. M¢ is called the quadratic approximation of M near p. It is an ana-logue for surfaces of a Frenet approximation of a curve.

From Definition 2.2 through Corollary 2.6 we have been concerned withthe geometry of M Ã R3 near one of its points p. These results thus applysimultaneously to all the points of the oriented region O on which, by ourinitial assumption, the unit normal U is defined. In particular then, we haveactually defined principal curvature functions k1 and k2 on O, where at eachpoint p of O, k1(p) and k2(p) are the principal curvatures of M at p. Note thatthese functions are only defined “modulo sign”: If U is replaced by -U, theybecome -k1 and -k2.

¢ = +( )M z k x k y:12 1

22

2

S U f fxy yyu u uu20

10

22( ) = -— = + .

S U f fxx xyu u uu10

10

21( ) = -— = + ,

u u1 21 0 0 0 1 0= ( ) = ( ), , and , ,

f x y f x f xy f yxx xy yy,( ) + +( )~ .12

20 2 0 0 2

FIG. 5.14

Exercises

1. Use the results of Example 1.3 to find the principal curvatures and prin-cipal vectors of

(a) The cylinder, at every point.(b) The saddle surface, at the origin.

2. If v π 0 is a tangent vector (not necessarily of unit length), show that thenormal curvature of M in the direction of v is

3. For each integer n � 2, let an be the curve t Æ (rcos t, rsin t, ±tn) in thecylinder M: x2 + y2 = r2. These curves all have the same velocity at t = 0;test Lemma 2.1 by showing that they all have the same normal componentof acceleration at t = 0.

4. For each of the following surfaces, find the quadratic approximation nearthe origin:

(a) z = exp (x2 + y2) - 1. (b) z = logcosx - log cosy.(c) z = (x + 3y)3.

5.3 Gaussian Curvature

The preceding section found the geometrical meaning of the eigenvalues andeigenvectors of the shape operator. Now we examine the determinant andtrace of S.

3.1 Definition The Gaussian curvature of M Ã R3 is the real-valued func-tion K = detS on M. Explicitly, for each point p of M, the Gaussian curva-ture K(p) of M at p is the determinant of the shape operator S of M at p.

The mean curvature of M Ã R3 is the function trace S. Gaussianand mean curvature are expressed in terms of principal curvature by

3.2 Lemma .

Proof. The determinant (and trace) of a linear operator may be definedas the common value of the determinant (and trace) of all its matrices. Ife1 and e2 are principal vectors at a point p, then by Theorem 2.5, we haveS(e1) = k1(p)e1 and S(e2) = k2(p)e2. Thus the matrix of S at p with respectto e1, e2 is

K k k H k k= = +( )1 2 1 2

12

,

H = 1 2/

k Sv v v v v( ) = ( ) • • .


5.3 Gaussian Curvature 217

This immediately gives the required result. ◆

A significant fact about the Gaussian curvature: It is independent of thechoice of the unit normal U. If U is changed to -U, then the signs of bothk1 and k2 change, so K = k1k2 is unaffected. This is obviously not the case withmean curvature which has the same ambiguity of sign asthe principal curvatures themselves.

The normal section method in Section 2 lets us tell, by inspection, approx-imately what the principal curvatures of M are at each point. Thus we get areasonable idea of what the Gaussian curvature K = k1k2 is at each point pby merely looking at the surface M. In particular, we can usually tell what thesign of K(p) is—and this sign has an important geometric meaning, whichwe now illustrate.

3.3 Remark The sign of Gaussian curvature at a point p.

(1) Positive. If K(p) > 0, then by Lemma 3.2, the principal curvatures k1(p)and k2(p) have the same sign. By Corollary 2.6, either k(u) > 0 for all unitvectors u at p or k(u) < 0. Thus M is bending away from its tangent plane Tp(M)in all tangent directions at p (Fig. 5.15)

The quadratic approximation of M near p is the paraboloid

(2) Negative. If K(p) < 0, then by Lemma 3.2, the principal curvatures k1(p)and k2(p) have opposite signs. Thus the quadratic approximation of M nearp is a hyperboloid, so M is also saddle-shaped near p (Fig. 5.16).

z k x k y= ( ) + ( )12

121

22

2p p .

H k k= +( )1 2 2 ,

k

k1

2

0

0

p

p

( )( )

ÊËÁ

ˆ¯.

FIG. 5.15 FIG. 5.16


(3) Zero. If K(p) = 0, then by Lemma 3.2 there are two cases:(a) If only one principal curvature is zero, say

(b) If both principal curvatures are zero, say

In case (a) the quadratic approximation is the cylinder so Mis trough-shaped near p (Fig. 5.17).

In case (b), the quadratic approximation reduces simply to the plane z = 0, so we get no information about the shape of M near p.

A torus of revolution T provides a good example of these different cases.At points on the outer half O of T, the torus bends away from its tangentplane as one can see from Fig. 5.18; hence K > 0 on O. But near each pointp of the inner half I, T is saddle-shaped and cuts through Tp(M). Hence K < 0 on I.

Near each point on the two circles (top and bottom) that separate O andI, the torus is trough-shaped; hence K = 0 there. (A quantitative check ofthese qualitative results is given in Section 7.)

In case 3(b) above, where both principal curvatures vanish, p is called aplanar point of M. (There are no planar points on the torus.) For example,the central point p of a monkey saddle, say

is planar. Here three hills and valleys meet, as shown in Fig. 5.19. Thus pmust be a planar point—the shape of M near p is too complicated for theother three possibilities in Remark 3.3.

We consider now some ways to compute Gaussian and mean curvature.

M z x x y x y: = +( ) -( )3 3 ,

2 12z k x= ( )p ,

k k1 2 0p p( ) = ( ) = .

k k1 20 0p p( ) π ( ) =, .

FIG. 5.17 FIG. 5.18


3.4 Lemma If v and w are linearly independent tangent vectors at a pointp of M Ã R3, then

Proof. Since v, w is a basis for the tangent plane Tp(M), we can write

This shows that

is the matrix of S with respect to the basis v, w. Hence

Using standard properties of the cross product, we compute

and a similar calculation gives the formula for H(p). ◆

Thus if V and W are tangent vector fields that are linearly independent ateach point of an oriented region, we have vector field equations

S V S W KV W

S V W V S W HV W

( ) ¥ ( ) = ¥

( ) ¥ + ¥ ( ) = ¥

,

2 .

S S a b c d

ad bc K

v w v w v w

v w p v w

( ) ¥ ( ) = +( ) ¥ +( )= -( ) ¥ = ( ) ¥ ,

K S ad bc H S a dp p( ) = = - ( ) = = +( )det ., trace12

12

a b

c dÊËÁ

ˆ¯

S a b

S c d

v v w

w v w

( ) = +

( ) = +

,

.

S S K

S S H

v w p v w

v w v w p v w

( ) ¥ ( ) = ( ) ¥

( ) ¥ + ¥ ( ) = ( ) ¥

,

2 .

FIG. 5.19

These may be solved for K and H by dotting each side with the normalvector field V ¥ W, and using the Lagrange identity (Exercise 6). We then find

The denominators are never zero, since the independence of V and W isequivalent to (V ¥ W) • (V ¥ W) > 0. In particular, the functions K and Hare differentiable.

Once K and H are known, it is a simple matter to find k1 and k2.

3.5 Corollary On an oriented region O in M, the principal curvaturefunctions are

Proof. To verify the formula, it suffices to substitute

and note that

◆

A more enlightening derivation (Exercise 4) uses the characteristic polyno-mial of S.

This formula shows only that k1 and k2 are continuous functions on O; theyneed not be differentiable since the square-root function is badly behaved atzero. The identity in the proof shows that H 2 - K is zero only at umbilic points,however, so k1 and k2 are differentiable on any oriented region free of umbilics.

A natural way to single out special types of surfaces in R3 is by restric-tions on Gaussian and mean curvature.

3.6 Definition A surface M in R3 is flat provided its Gaussian curvatureis zero, and minimal provided its mean curvature is zero.

H Kk k

k kk k2 1 2

2

1 21 2

2

4 4- =

+( )- =

-( ).

K k k Hk k

= =+

1 21 2

2and

k k H H K1 22, = ± - .

K

S V V S V W

S W V S W VV V V W

W V W W

H

S V V S V W

W V W W

V V V W

S W V S W WV V V W

W V W W

=

( ) ( )( ) ( )

=

( ) ( )+ ( ) ( )

• •

• •• •

• •

• •

• •

• •

• •• •

• •

.

,

2


As expected, a plane is flat, for by Example 1.3 its shape operators are allzero, so K = det S = 0. On a circular cylinder, (3) of Example 1.3 shows thatS is singular at each point p, that is, has rank less than the dimension of thetangent plane Tp(M). Thus, although S itself is never zero, its determinant isalways zero, so cylinders are also flat. This terminology seems odd at first fora surface so obviously curved, but it will be amply justified in later work.

Note that minimal surfaces have Gaussian curvature K � 0, because if

then k1 = -k2, so K = k1k2 � 0.Another notable class of surfaces consists of those with constant Gauss-

ian curvature. As mentioned earlier, Example 1.3 shows that a sphere ofradius r has (for U outward). Thus the sphere S has constantpositive curvature : The smaller the sphere, the larger its curvature.

We shall find many examples of these various special types of surface as we proceed through this chapter.

Exercises

1. Show that there are no umbilics on a surface with K < 0, and that whenK � 0, umbilic points are planar.

2. Let u1 and u2 be orthonormal tangent vectors at a point p of M. Whatgeometric information can be deduced from each of the following conditionson S at p?

(a) S(u1) • u2 = 0. (b) S(u1) + S(u2) = 0.(c) S(u1) ¥ S(u2) = 0. (d) S(u1) • S(u2) = 0.

3. (Mean curvature.) Prove that(a) the average value of the normal curvature in any two orthogonal direc-tions at p is H(p). (The analogue for K is false.)

(b)

where k(J) is normal curvature.

4. The characteristic polynomial of an arbitrary linear operator S is

where A is any matrix of S.

p k A kI( ) = -( )det ,

H k dp( ) = ( ) ( )Ú1 20

2

/ p J Jp

,

K r= 1 2

k k r1 2 1= = -

Hk k

=+

=1 2

20,


(a) Show that the characteristic polynomial of the shape operator is

(b) Every linear operator satisfies its characteristic equation; that is, p(S)is the zero operator when S is formally substituted in p(k). Prove this inthe case of the shape operator by showing that

for any pair of tangent vectors to M.The real-valued functions

defined for all pairs of tangent vectors to an oriented surface, are tradition-ally called the first, second, and third fundamental forms of M. They are notdifferential forms; in fact, they are symmetric in v and w rather than alter-nating. The shape operator does not appear explicitly in the classical treat-ment of this subject; it is replaced by the second fundamental form.

5. (Dupin curve.) For a point p of an oriented region of M, let C0 be theintersection of M near p with its tangent plane ; specifically, C0 con-sists of those points of M near p that lie in the plane through p orthogonalto U(p). C0 may be approximated by substituting for M its quadratic approx-imation M; thus C0 is approximated by the curve

(a) Describe C0 in each of the three cases K(p) > 0, K(p) < 0, and K(p) = 0(not planar).(b) Repeat (a) with C0 replaced by Ce and C-e, where the tangent plane hasbeen replaced by the two parallel planes at distance ±e from it.(c) This scheme fails for planar points since the quadratic approximationbecomes M: z = 0. For the monkey saddle, sketch C0, Ce, and C-e.

6. For vectors x, y, v, w in R3, prove the Lagrange identity

(a) By hand. (Hint: Since both sides are linear in each vector separately, itsuffices to prove the identity when each vector is one of the unit vectorsu1, u2, u3.)(b) By computer. (For dot and cross products, see the Appendix.)

x y v wx v x w

y v y w¥( ) ¥( ) =•

• •

• •.

ˆ : .C k x k y0 12

22 0 0 0+ = ( ), near ,

T Mp ( )

I , , II , ,

III , ,

v w v w v w v w

v w v w v w

( ) = ( ) = ( )( ) = ( ) = ( ) ( )

• •

• •

S

S S S2

S S HS Kv w v w v w( ) ( ) - ( ) + =• • •2 0

k Hk K2 2- + .



7. (Parallel surfaces.) Let M be a surface oriented by U; for a fixed numbere (positive or negative) let F: M Æ R3 be the mapping such that

(a) If v is tangent to M at p, show that = F*(v) is v - eS(v). Deduce that

where

When the function J does not vanish on M (for example, if M is compactand |e | small), this shows that F is a regular mapping, so the image

is at least an immersed surface in R3 (Ex. 16 in Sec. 4.8). is said to be par-allel to M at distance e (Fig. 5.20).

(b) Show that the canonical isomorphisms of R3 make U a unit normal onfor which .

(c) Derive the following formulas for the Gaussian and mean curvaturesof M:

8. (Continuation.)(a) Check the results in (c) in the case of a sphere of radius r oriented bythe outward normal U. Describe the mapping F = Fe when e is 0, -r, and-2r.(b) Starting from an orientable surface with constant positive Gaussiancurvature, construct a surface with constant mean curvature.

K FKJ

H FH K

J( ) = ( ) =

-; .

e

S Sv v( ) = ( )M

M

M F M= ( )

J H K k k= - + = -( ) -( )1 2 1 121 2e e e e .

v w p v w¥ = ( ) ¥J ,

v

F Up p p( ) = + ( )e .

FIG. 5.20


5.4 Computational Techniques

We have defined the shape operators S of a surface M in R3 and found geo-metrical meaning for its main algebraic invariants: Gaussian curvature K,mean curvature H, principal curvatures k1 and k2, and (at each point) prin-cipal vectors e1 and e2. We shall now see how to express these invariants interms of patches in M.

If x: D Æ M is a patch in M Ã R3, we have already used the three real-valued functions

on D. Here E > 0 and G > 0 are the squares of the speeds of the u- and v-parameter curves of x, and F measures the coordinate angle J between xu andxv, since

(Fig. 5.21). E, F, and G are the “warping functions” of the patch x: Theymeasure the way x distorts the flat region D in R2 in order to apply it to thecurved region x(D) in M. These functions completely determine the dotproduct of tangent vectors at points of x(D), for if

then

(In such equations we understand that xu, xv, E, F, and G are evaluated at (u, v) where x(u, v) is the point of application of v and w.)

Now xu ¥ xv is a function on D whose value at each point (u, v) of D is avector orthogonal to both xu(u, v) and xv(u, v)—and hence normal to M atthe point x(u, v). Furthermore, by Exercise 6 of Section 3,

x xu v EG F¥ = -2 2 .

v w• .= + +( ) +Ev w F v w v w Gv w1 1 1 2 2 1 2 2

v x x w x x= + = +v v w wu v u u1 2 1 2and ,

F EGu v u v= = =x x x x• cos cosJ J

E F Gu u u v v u v v= = = =x x x x x x x x• • • •, ,

FIG. 5.21

Since x is, by definition, regular, this real-valued function on D is never zero.Thus we can construct the unit normal function

on D, which assigns to each (u, v) in D a unit normal vector to M at x(u, v).We emphasize that in this context, U, like xu and xv, is not a vector field onx(D), but merely a vector-valued function on D. Nevertheless, we may regardthe system xu, xv, U as a kind of defective frame field. At least U has unitlength and is orthogonal to both xu and xv, even though xu and xv are gener-ally not orthonormal.

In this context, covariant derivatives are usually computed along the para-meter curves of x, where by the discussion in Section 1, they reduce to partialdifferentiation with respect to u and v. As in the case of xu and xv, these partialderivatives are again denoted by subscripts u and v. If

then just as for xu and xv on page 140, we have

Evidently xuu and xvv give the accelerations of the u- and v-parametercurves. Since order of partial differentiation is immaterial, xuv = xvu, whichgives both the covariant derivative of xu in the xv direction and of xv in thexu direction.

Now if S is the shape operator derived from U, we define three more real-valued functions on D:

Because xu, xv gives a basis for the tangent space of M at each point ofx(D), it is clear that these functions uniquely determine the shape operator.Since this basis is generally not orthonormal, , , and do not lead to simple

L

M

N

,

,

= ( )= ( ) = ( )= ( )

S

S S

S

u u

u v v u

v v

x x

x x x x

x x

•

• •

• .

x

x

x

x

x

x

uu

vu

vv

xu

xu

xu

xu v

xu v

xu v

xv

xv

xv

=∂∂

∂∂

∂∂

ÊËÁ

ˆ¯

=∂

∂ ∂∂

∂ ∂∂

∂ ∂ÊËÁ

ˆ¯

=∂∂

∂∂

∂∂

ÊËÁ

ˆ¯

21

2

22

2

23

2

21

22

23

21

2

22

2

23

2

, , ,

, , ,

, , .

x u v x u v x u v x u v, , , , , , ,( ) = ( ) ( ) ( )( )1 2 3

U u v

u v

=¥¥

x xx x

5.4 Computational Techniques 225

expressions for S(xu) and S(xv) in terms of xu and xv. In the formulas preceding Corollary 3.5, however, they do provide simple expressions for Gaussian and mean curvature.

4.1 Corollary If x is a patch in M Ã R3, then

Proof. Evaluated on x(D), the formulas on page 220 express K and H interms of tangent vector fields V and W. If the latter are replaced by xu andxv, respectively, we find the required formulas for K and H. ◆

When the patch x is clear from context, we shall usually abbreviate thecomposite functions K(x) and H(x) to merely K and H.

By a device like that used in Lemma 2.1, we can find a simple way tocompute , , and —and thereby K and H. For example, since U • xu = 0,partial differentiation with respect to v—that is, ordinary differentiationalong v-parameter curves—yields

(Recall that Uv is the covariant derivative of the vector field v Æ U (u0, v) oneach v-parameter curve u = u0.) Since xv gives the velocity vectors of suchcurves, Exercise 1.1 shows that Uv = -S(xv). Thus the preceding equationbecomes

(Fig. 5.22). Three similar equations may be found by replacing u by v, and vby u. In particular,

Again, since xu and xv give a basis for the tangent space at each point, this issufficient to prove that S is symmetric (Lemma 1.4).

4.2 Lemma If x is a patch in M Ã R3, then

L

M

N

,

,

= ( ) =

= ( ) =

= ( ) =

S U

S U

S U

u u uu

u v uv

v v vv

x x x

x x x

x x x

• •

• •

• • .

S U U Su v vu uv v ux x x x x x( ) = = = ( )• • • • .

S Uv u uvx x x( ) =• •

0 =∂

∂( ) = +

vU U Uu v u uv• • • .x x x

KEG F

HG E F

EG Fx x( ) =

--

( ) =+ -

-( )LN M L N M

,2

2 2

22

.



The first equation in each case is just the definition, and u and v may bereversed in the formulas for .

4.3 Example Computation of Gaussian and mean curvature(1) Helicoid (Exercise 5 of Section 4.2). This surface H, shown in Fig. 5.23,

is covered by a single patch

x u v u v u v bv b, , , , ,( ) = ( ) πcos sin 0

FIG. 5.22

FIG. 5.23

for which

Hence

Because coordinate patches are, by definition, regular mappings, we have seenin Chapter 4 that the function

is never zero. For any patch we denote this useful function by W, that is,

In the case at hand, so the unit normal function is

(A computation of U can always be checked by verifying that the result is aunit vector orthogonal to both xu and xv.)

Next we find

Here xuu = 0 is obvious, since the u-parameter curves are straight lines. Thev-parameter curves are helices, and this formula for the acceleration xvv wasfound already in Chapter 2. Now by Lemma 4.2,

Hence by Corollary 4.1 and the results above,

L

M

N

,

,

=¥( )

=

=¥( )

= -

=¥( )

=

xx x

xx x

xx x

uuu v

uvu v

vvu v

W

Wb

W

W

•

•

• .

0

0

x

x

x

uu

uv

vv

v v

u v u v

=

= -( )= - -( )

0

0

0

,

, , ,

, ,

sin cos

cos sin .

UW

b v b v u

b u

u v=¥

=-( )

+x x sin cos

., ,

2 2

W W u v b u= ( ) = +, ,2 2

W EG F= - 2 .

x xu v EG F¥ = - 2

x xu v b v b v u¥ = -( )sin cos ., ,

x

xu

v

v v E

u v u v b F

G b u

= ( ) == -( ) =

= +

cos sin

sin cos

.

, , , ,

, , , ,

0 1

02 2


Thus the helicoid is a minimal surface with Gaussian curvature

The minimum value occurs on the central axis (u = 0) of the helicoid, and K Æ 0 as distance |u| from the axis increases to infinity.

(2) The saddle surface M: z = xy (Example 1.3). This time we use theMonge patch x(u, v) = (u, v, uv) and with the same format as above, compute

Hence

Strictly speaking, these functions are K(x) and H(x) defined on the domainR2 of x. In this case, it is easy to express K and H directly as functions on M by using the cylindrical coordinate functions and z. Notefrom Fig. 5.24 that

and

hence on M,

Kr

Hz

r=

-+( )

=-

+( )1

1 12 2 2 3 2, .

z u v uvx ,( )( ) = ;

r u v u vx ,( )( ) = +2 2

r x y= +2 2

Ku v

Huv

u v=

-+ +( )

=-

+ +( )1

1 12 2 2 2 2 3 2, .

x

x

x

x

x

u

v

uu

uv

vv

v E v

u F uv

G u

U v u W W u v

W

= ( ) = += ( ) =

= +

= - -( ) = + += == ( ) == =

1 0 1

0 1

1

1 1

0 0

0 0 1 1

0 0

2

2

2 2

, , ,

, , , ,

,

, , ,

, ,

, , , ,

,

L

N

M

,

,

.

K b= -1 2

- £ <1

02b

K .

KEG F

b WW

bW

b

b u

HG E F

EG F

=--

=-( )

=-

=-+( )

=+ -

-( ) =

LN M

L N M

,2

2

2

2

2

4

2

2 2 2

2

22

0.



Thus the Gaussian curvature of M depends only on distance to the z axis,rising from K = -1 (at the origin) toward zero as r goes to infinity, while Hvaries more radically.

Like all simple (that is, one-patch) surfaces, the helicoid and saddle sur-faces are orientable, since computations as above provide a unit normal onthe whole surface. Thus the principal curvature functions k1 ≥ k2 can bedefined unambiguously on each surface. These can always be found from Kand H by Corollary 3.5. Since the helicoid is a minimal surface, we get thesimple result

For the saddle surface,

Techniques for computing principal vectors are left to the exercises.The computational results in this section, though stated for coordinate

patches, remain valid for arbitrary regular mappings x: D Æ R3 since therestriction of x to any small enough open set in D is a patch.

Exercises

1. For the geographical parametrizationof the sphere S of radius r, find E,x u v r v u r v u r v, , ,( ) = ( )cos cos cos sin sin ,

k kz r z

r1 2

2 2

2 3 2

1

1, =

- ± + ++( )

.

k kb

b u1 2 2 2, =

±+( ) .

FIG. 5.24

F, G, and W, and then the unit normal U, Gaussian curvature K, and meancurvature H.

2. For a Monge patch x(u, v) = (u, v, f(u, v)), show that,

where

Then find formulas for K and H.

3. (Continuation.) Deduce that the image of x is(a) flat if and only if

(b) minimal if and only if

4. Let x be the patch

defined on Show that the image of x is a minimal surfacewith Gaussian curvature

where W 2 = 1 + tan2u + tan2v. (This patch is in Scherk’s surface, Ex. 5 ofSec. 5.5.)

5. Show that a curve segment

has length

where E, F, and G are evaluated on a1, a2.

L Ea Fa a Ga dta

b

a( ) = ¢ + ¢ ¢ + ¢( )Ú 12

1 2 22

12

2 ,

a t a t t a t ba( ) = ( ) ( )( ) £ £x 1 2, , ,

Ku v

W=

- sec sec2 2

4,

- < <p p/ , / .2 2u v

x u v u v v u, , ,( ) = -( )log cos log cos

1 2 1 02 2+( ) - + +( ) =f f f f f f fu vv u v uv v uu .

f f fuu vv uv- =2 0;

W EG F f fu v= - = + +( )2 2 2 1 21

E ffW

F f ffW

G ff

W

uuu

u vuv

vvv

= + =

= =

= + =

1

1

2

2

, ,

, ,

,

L

M

N ,


6. Find the Gaussian curvature of the elliptic and hyperbolic paraboloids

where e = ±1.

7. Find the curvature of the monkey saddle M: z = x3 - 3xy2, and expressit in terms of

8. A patch x in M is orthogonal provided F = 0 (so xu and xv are orthogo-nal at each point). Show that in this case

(b) A patch x in M is principal provided F = = 0. Prove that xu and xv

are principal vectors at each point, with corresponding principal curvatures/E and /G.

9. Prove that a nonzero tangent vector v = v1xu + v2xv is a principal vectorif and only if

(Hint: v is principal if and only if S(v) ¥ v = 0.)

10. Show that on the saddle surface z = xy the two vector fields

are principal at each point. Check that they are orthogonal and tangent to M.

11. If v = v1xu + v2xv is tangent to M at x(u, v), show that the normal

curvature in the direction is

where the various functions are evaluated at (u, v).

kv v v v

Ev Fv v Gvu( ) =

+ ++ +

12

1 2 22

12

1 2 22

2

2

M N,

u v v= /

1 1 1 12 2 2 2+ ± + + ± +( )x y y x x y,

v v v v

E F G22

1 2 12

0

-=

L M N

.

SE G

SE G

u u v

v u v

x x x

x x x

( ) = +

( ) = +

L M

M N

,

.

r x y= +2 2 .

M zxa

yb

: = +2

2

2

2e ,



12. Show that a ruled surface x(u, v) = b(u) + vd(u) has Gaussian curvature

where W = ||b ¢ ¥ d + vd¢ ¥ d ||.

13. (Flat ruled surfaces.)(a) Show that generalized cones and cylinders are flat (Exs. 3 and 4 of Sec.4.2).(b) If b is a unit-speed curve in R3 with k > 0, the ruled surface

where T(u) = b¢(u), is called the tangent surface of b. Prove that x is regularand the tangent surface is flat. (The surface is separated into two pieces bythe curve; Fig. 5.25 shows the v > 0 half.)

14. (Patch criterion for umbilics.)(a) Show that a point x(u, v) is umbilic if and only if there is a number ksuch that at (u, v),

(Then k is the principal curvature k1 = k2.)

15. Find the umbilic points, if any, on the following surfaces:(a) Monkey saddle (Ex. 7).(b) Elliptic paraboloid (Ex. 6), assuming a � b.

(Hint: Compute the “vectors” (E, F, G) and (, , ) for arbitrary (u, v), dis-carding common factors if convenient. Then solve (E, F, G) ¥ (, , ) = 0for (u, v).)

L M N, ,= = =kE kF kG .

x u v u vT u v, , ,( ) = ( ) + ( ) πb 0

KEG F W

=-

-=

- ¢ ¥ ¢( )M,

2

2

2

4

b d d•

FIG. 5.25

16. (Loxodromes.) For a π 0, let fa: be the unique func-tion such that

If x is the geographical parametrization of a sphere, the curve la(t) = x(fa(t),t)is a loxodrome.

(a) Prove that la¢ always makes a constant angle with the due-north vectorfield xv. Thus la represents a trip with constant (idealized) compassbearing.(b) Show that the length of la from the south pole (0, 0, -r) to the north pole (0, 0, r) (limit values) is .

(c) (Computer.) Verify that fa(t) = a logtan , and plot l10 fromnear the south pole to near the north pole on a unit sphere. (Requiresmoothness, and keep the same scale on each axis.)

17. (Tubes.) If b is a curve in R3 with 0 < k � b, let

Thus the v-parameter curves are circles of constant radius e in planes orthog-onal to b. Show that

(a) xu ¥ xv = -e(1 - ke cos v)(cos vN(u) + sin vB(u)).(b) If e is small enough, x is regular. So x is at least an immersed surface,called a tube around b.(c) U = cos v N(u) + sin v B(u) is a unit normal vector on the tube.

(d)

(Hint: Use S(xu) ¥ S(xv) = K xu ¥ xv.)

The following exercises deal with use of the computer in patch com-putations.

18. (Computer.) The Appendix gives computer commands for the functionsE, F, G, W, , , derived from a patch.

(a) Write the computer commands, based on Corollary 4.1, that give theGaussian curvature and mean curvature of a patch in terms of these functions.(b) To test these commands, find E, F, G, W, , , , K, H for each of thecases in Example 4.3. Compare with the text computations.

19. (Computer.) Make a save file containing the following patches orparametrizations. (See Appendix for “save files” and format for parameters.)

(a) the patch in Exercise 4.

Ku vu v

=- ( )- ( )( )k

e k ecos

cos.

1

x u v u v N u v B u,( ) = ( ) + ( ) + ( )( )b e cos sin .

t / /2 4+( )p

1 2+ a rp

¢( ) = ( ) =f ta

tfa cos

.and 0 0

-( ) Æp p/ , /2 2 R


(b) a single Monge patch—with parameters a, b, e—for the hyperboloidsin Exercise 6.(c) a Monge patch for the monkey saddle (Ex. 7), in terms of (i) rectan-gular coordinates u, v and (ii) polar coordinates r, J on R2.(d) the parametrization of Enneper’s surface in Exercise 15 of Section 7.(e) the geographical parametrization of a sphere of radius r.

20. (Computer formulas.)(a) For a patch x in R3, show that Gaussian curvature can be expresseddirectly in terms of x as

This formula gives the fastest general computer computation of K. TheAppendix has computer commands for it in the Mathematica and Maplesystems.

(b) Test this command on the two cases in Example 4.3 and the patches inExercise 19.

The derivation of the corresponding formula for mean curvature is rathertedious. This formula may be found in Alfred Gray’s book [G].(c) Find a computer formula for the Gaussian curvature of the graph ofa function f: R2 Æ R. (Hint: use Ex. 2.) Test this command on the Mongepatches referenced in Exercises 18 and 19.

21. (Computer.)(a) Write commands that, given a curve a on some interval, plot the tubeof radius r around a. (See Ex. 17.)(b) Use part (a) to plot the tube of radius around two turns of the helix

t Æ (3 cos t, 3 sin t, ).(c) Plot the tube of radius around the curve t in Exercise 19 of Sec. 2.4.(This makes it clear that t is a trefoil knot.)

5.5 The Implicit Case

In this brief section we describe a way to compute the geometry of a surfaceM Ã R3 that has a nonvanishing normal vector field Z defined on the entiresurface. The main case is a surface given in implicit form M: g = 0, for then,by Lemma 3.8 of Chapter 4, the gradient

is such a vector field.

— =∂∂Âg

gxi

12

t 2

12

K u v uu u v vv u v uv u v

u u v v u v

,( ) =¥( ) ¥( ) - ¥( )

( )( ) - ( )( )x x x x x x x x x

x x x x x x

• • •

• • •.

2

2 2

5.5 The Implicit Case 235


Let S be the shape operator on M derived from the unit normal

Write Then if V is a tangent vector field on M, Method 2 inSection 1 gives

Hence, using a Leibnizian property of such derivatives,

(Fig. 5.26). The last term here, V[1/||Z||] Z, is evidently a normal vector field;we do not care which one it is, so we denote it merely by -NV. Thus

Note that if W is another tangent vector field on M, then NV ¥ NW = 0, whileproducts such as NV ¥ Y are tangent to M for any Euclidean vector field Yon M. Thus it is a routine matter to deduce the following lemma from Lemma3.4.

5.1 Lemma Let Z be a nonvanishing normal vector field on M. If V andW are tangent vector fields such that V ¥ W = Z, then

KZ Z

Z

H ZZ W V Z

Z

V W

V W

=— ¥ —

= -— ¥ + ¥ —

Z •

• .

4

32

,

S V UZ

ZNV

VV( ) = -— = -

—+ .

— = — ÊËÁ

ˆ¯

=—

+ ÈÎÍ

˘˚

V VVU

ZZ

ZZ

VZ

Z1

— = [ ]ÂV i iZ v z U .

Z zUi i= Â .

U Z Z= .

V

V

FIG. 5.26

To compute, say, the Gaussian curvature of a surface M: g = c usingpatches, one must begin by explicitly finding enough of them to cover all ofM; a complete computation of K may thus be tedious, even when g is a rathersimple function. The following example shows to advantage the approach justdescribed.

5.2 Example Curvature of the ellipsoid

We write g = , and use the (nonvanishing) normal vector field

Then if is a tangent vector field on M,

since

Similar results for another tangent vector field W yield

where X is the special vector field used in Example 3.9 of Chapter 4.It is always possible to choose V and W so that V ¥ W = Z. But then

Thus by Lemma 5.1 we have found

that is,

Ka a a Z

Zxa

i

i

= = ÊËÁ

ˆ¯Â1

12

22

32 4

42

4

2

, where ,

X V W X Zxa

i

i

• • .¥ = = =Â2

21

xUi iÂ

Z Z Z

xa

xa

xa

va

va

va

wa

wa

wa

a a aX V WV W• •— ¥ — = = ¥

1

12

2

22

3

32

1

12

2

22

3

32

1

12

2

22

3

32

12

22

32

1,

V x dx V vi i i[ ] = ( ) = .

— = [ ] =Â ÂVi

ii

i

iiZ

V xa

Uva

U2 2

,

V vUi i= Â

Z gxa

Ui

ii= — = Â1

2 2.

xa

i

i

2

2Â

M gxa

yb

zc

: .= + + =2

2

2

2

2

21



For any oriented surface in R3, its support function h assigns to each pointp the orthogonal distance h(p) = p • U(p) from the origin to the Euclideantangent plane , as shown in Fig. 5.27 for the ellipsoid. Using the above-mentioned vector field X (whose value at p is the tangent vector pp), we findfor the ellipsoid that

Thus a clearer expression of the Gaussian curvature of the ellipsoid is

Note that if a = b = c = r (so M is a sphere), then has constantvalue r, and this formula reduces to

Exercises

1. Show that the elliptic hyperboloids of one and two sheets (Ex. 2.9 of Ch.4) have Gaussian curvatures respec-tively, where both support functions h are given by the same formula as forthe ellipsoid in Example 5.2.

2. If h is the support function of an oriented surface M Ã R3, show that(a) A point p of M is a critical point of h if and only if p • S(v) = 0 for all tangent vectors to M at p. (Hint: Write h as X • U, where )X xUi i= Â .

K h a b c K h a b c= - =4 2 2 2 4 2 2 2/ / ,and

K r= 1 2/ .h Z= 1

Kh

a b c=

4

2 2 2.

h X U XZZ Z

= = =• • .1

T Mp ( )

Ka b c Z

Zxa

yb

zc

= = + +ÊËÁ

ˆ¯

12 2 2 4

42

4

2

4

2

4

2

, .where

FIG. 5.27


(b) When K(p) π 0, p is a critical point of h if and only if p (consideredas a vector) is orthogonal to M at p.

3. (a) Use the preceding exercises to find the critical points of the Gauss-ian curvature function K on the ellipsoid and on the hyperboloids of one andtwo sheets (Ex. 2.9 of Ch. 4).

(b) Assuming a � b � c for these surfaces, find their Gaussian curvatureintervals.

4. Compute K and H for the saddle surface M: z = xy by the method ofthis section. (Hint: Take V and W tangent to the rulings of M.)

5. Scherk’s minimal surface, M: ezcosx = cos y. Let R be the region in thexy plane on which cosx cosy > 0. R is a checkerboard pattern of open squares,with vertices Show that:

(a) M is a surface.(b) For each point (u, v) in R there is exactly one point (u, v, w) in M. Theonly other points of M are entire vertical lines over each of the vertices ofR (Fig. 5.28).(c) M is a minimal surface with (Hint: V = cosx U1 + sinx U3 is a tangent vector field.)(d) The patch in Exercise 4 of Section 4 parametrizes the part of M overa typical open square. Show that the curvature K(u, v) calculated there isconsistent with (c).

K e e xz z= - +( )2 2 2 21/ sin .

p p p p/ , / .2 2+ +( )m n

FIG. 5.28

6. Let Z be a nonvanishing normal vector field on M. Show that a tangentvector v to M at p is principal if and only if

(Hint: Recall that v is principal if and only if S(v) ¥ v = 0.)The preceding equation together with the tangency equation Z(p) • v = 0

can be solved for the principal directions. Thus umbilics can be located usingthese equations, since p is umbilic if and only if every tangent vector at p isprincipal.

7. For the ellipsoid M: , show that:(a) A tangent vector v at p is principal if and only if

(b) Assuming a1 > a2 > a3, there are exactly four umbilics on M, with coordinates

5.6 Special Curves in a Surface

In this section we consider three geometrically significant types of curves ina surface M Ã R3.

6.1 Definition A regular curve a in M Ã R3 is a principal curve providedthat the velocity a ¢ of a always points in a principal direction.

Thus principal curves always travel in directions for which the bending ofM in R3 takes its extreme values. Neglecting changes of parametrization,there are exactly two principal curves through each nonumbilic point of M—and these necessarily cut orthogonally across each other. (At an umbilic pointp, every direction is principal, and near p the pattern of principal curves canbe quite complicated.)

6.2 Lemma Let a be a regular curve in M Ã R3, and let U be a unitnormal vector field restricted to a. Then

(1) The curve a is principal if and only if U¢ and a ¢ are collinear at eachpoint.

p aa aa a

p p aa aa a1 1

12

22

12

32

1 2

2 3 322

32

12

32

1 2

0= ±--

ÊËÁ

ˆ¯ = = ±

--

ÊËÁ

ˆ¯, , .

0 1 2 3 22

32

2 3 1 32

12

3 1 2 12

22= -( ) + -( ) + -( )p v v a a p v v a a p v v a a .

x ai i2 2 1Â =

v p• .Z Zv( ) ¥ — = 0


5.6 Special Curves in a Surface 241

(2) If a is a principal curve, then the principal curvature of M in the direc-tion of a ¢ is

Proof. (1) Exercise 1.1 shows that S(a ¢) = -U¢. Thus U¢ and a ¢ arecollinear if and only if S(a ¢) and a ¢ are collinear. But by Theorem 2.5,this amounts to saying that a ¢ always points in a principal direction or,equivalently, that a is a principal curve.

(2) Since a is a principal curve, the vector field consists entirelyof (unit) principal vectors belonging to, say, the principal curvature ki. Thus

where the last equality uses Lemma 2.1. ◆

In this lemma, (1) is a simple criterion for a curve to be principal, while (2)gives the principal curvature along a curve known to be principal.

6.3 Lemma Let a be a curve cut from a surface M Ã R3 by a plane P. Ifthe angle between M and P is constant along a, then a is a principal curveof M.

Proof. Let U and V be unit normal vector fields to M and P (respec-tively) along the curve a, as shown in Fig. 5.29. Since P is a plane, V isparallel, that is, V¢ = 0. The constant-angle assumption means that U • Vis constant; thus

k k S

S U

i = ¢ ¢( ) = ¢ ¢( ) ¢ ¢

=¢( ) ¢

¢ ¢=

¢¢¢ ¢

a a a a a a

a aa a

aa a

/ / • /

••

••

,

¢ ¢a a/

¢¢ ¢ ¢a a a• / • .U

FIG. 5.29


Since U is a unit vector, U¢ is orthogonal to U as well as to V. The sameis of course true of a ¢, since a lies in both M and P. If U and V are lin-early independent (as in Fig. 5.29) we conclude that U¢ and a ¢ are collinear;hence by Lemma 6.2, a is principal.

However, linear independence fails only when U = ±V. But then U ¢ = 0,so a is (trivially) principal in this case as well. ◆

Using this result, it is easy to see that the meridians and parallels of a surfaceof revolution M are its principal curves. Indeed, each meridian m is sliced fromM by a plane through the axis of revolution and hence orthogonal to M alongm, while each parallel p is sliced from M by a plane orthogonal to the axis,and by rotational symmetry such a plane makes a constant angle with Malong p.

Directions tangent to M Ã R3 in which the normal curvature is zero arecalled asymptotic directions. Thus a tangent vector v is asymptotic providedk(v) = S(v) • v = 0, so in an asymptotic direction, M is (instantaneously, atleast) not bending away from its tangent plane.

Using Corollary 2.6 we can get a complete analysis of asymptotic direc-tions in terms of Gaussian curvature.

6.4 Lemma Let p be a point of M Ã R3.(1) If K(p) > 0, then there are no asymptotic directions at p.(2) If K(p) < 0, then there are exactly two asymptotic directions at p, and

these are bisected by the principal directions (Fig. 5.30) at angle J such that

(3) If K(p) = 0, then every direction is asymptotic if p is a planar point;otherwise there is exactly one asymptotic direction and it is also principal.

tan .2 1

2

J =- ( )

( )k

kp

p

0 = ( )¢ = ¢U V U V• • .

FIG. 5.30


Proof. These cases all derive from Euler’s formula

in Corollary 2.6.(1) Since k1(p) and k2(p) have the same sign, k(u) is never zero.(2) Here k1(p) and k2(p) have opposite signs, and we can solve the

equation 0 = k1(p) cos2J + k2(p) sin2J to obtain the two asymptoticdirections.

(3) If p is planar, then

hence k(u) is identically zero. If just k2(p) = 0, then

is zero only when cos J = 0, that is, in the principal direction u = e2. ◆

We can get an approximate idea of the asymptotic directions at a point pof a given surface M by picturing the intersection of the tangent plane with M near p. When K(p) is negative, this intersection will consist of twocurves through p whose tangent lines (at p) are asymptotic directions (Exer-cise 5 of Section 53).

Figure 5.31 shows the two asymptotic directions A and A¢ at a point p onthe inner equator of a torus.

T Mp ( )

k ku p( ) = ( )12cos J

k k1 2 0p p( ) = ( ) = ;

k k ku p p( ) = ( ) + ( )12

22cos sinJ J

FIG. 5.31

6.5 Definition A regular curve a in M Ã R3 is an asymptotic curve pro-vided its velocity a ¢ always points in an asymptotic direction.

Thus a is asymptotic if and only if

k S¢( ) = ¢( ) ¢ =a a a• .0

Since S(a ¢) = -U¢, this gives a criterion, U¢ • a ¢ = 0, for a to be asymptotic.Asymptotic curves are more sensitive to Gaussian curvature than are princi-pal curves: Lemma 6.3 shows that there are none in regions where K is pos-itive, but two cross (at an angle depending on K) at each point of a regionwhere K is negative.

The simplest criterion for a curve in M to be asymptotic is that its accel-eration a ≤ always be tangent to M. In fact, differentiation of U • a ¢ = 0 gives

so U¢ • a ¢ = 0 (a asymptotic) if and only if U • a ≤ = 0.The analysis of asymptotic directions in Lemma 6.4 has consequences for

both flat and minimal surfaces. First, a surface M in R3 is minimal if and onlyif there exist two orthogonal asymptotic directions at each of its points. In fact,H(p) = 0 is equivalent to k1(p) = -k2(p), and an examination of the possi-bilities in Lemma 5.4 shows that k1(p) = -k2(p) if and only if either (a) p isplanar (so the criterion holds trivially) or (b)

which means that the two asymptotic directions are orthogonal.Thus a surface is minimal if and only if through each point there are two

asymptotic curves that cross orthogonally. This observation gives geometricmeaning to the calculations in Example 4.3, which show that the helicoid isa minimal surface. In fact, the u- and v-parameter curves of the patch x areorthogonal since F = 0, and their accelerations are tangent to the surfacesince = U • xuu = 0 and = U • xvv = 0.

Recall that a ruled surface is swept out by a line moving through R3 (Def-inition 2.6 in Chapter 4). We have seen, for example, that the helicoid andsaddle surface in Example 4.3 are ruled surfaces. Thus it is no accident thatboth these surfaces have K negative, since:

6.6 Lemma A ruled surface M has Gaussian curvature K � 0. Further-more, K = 0 if and only if the unit normal U is parallel along each ruling of M(so all points p on a ruling have the same Euclidean tangent plane ).

Proof. A straight line t Æ p + tq is certainly asymptotic since its acceler-ation is zero and is thus trivially tangent to M. By definition a ruled surfacecontains a line through each of its points, so there is an asymptotic direc-tion at each point. Hence, by Lemma 6.4, K � 0.

Now let a(t) = p + tq be an arbitrary ruling in M. If U is parallel alonga, then S(a ¢) = -U¢ = 0. Thus a is a principal curve with principal curva-ture k(a ¢) = 0, so K = k1 k2 = 0.

T Mp ( )

K p( ) < = ±0 4with ,J p

¢ ¢ + ¢¢ =U U• •a a 0,


Conversely, if K = 0 we deduce from Case (3) in Lemma 6.4 that asymp-totic directions (and curves) in M are also principal. Thus each ruling isprincipal (S(a ¢) = k(a ¢)a ¢) as well as asymptotic (k(a ¢) = 0); hence

U¢ = -S(a ¢) = 0, ◆

and U is parallel along each ruling of M.

We come now to the last and most important of the three types of curvesunder discussion.

6.8 Definition A curve a in M Ã R3 is a geodesic of M provided its accel-eration a ≤ is always normal to M.

Since a ≤ is normal to M, the inhabitants of M perceive no acceleration atall—for them the geodesic is a “straight line.” A full study of geodesics isgiven in later chapters, where, in particular, we examine their character asshortest routes of travel. Geodesics are far more plentiful in a surface M thanare principal or asymptotic curves. Indeed, Theorem 4.2 of Chapter 7 willshow that given any tangent vector v to M there is a (unique) geodesic withinitial velocity v.

Because the acceleration a ≤ of a geodesic is orthogonal to M, it is orthog-onal to the velocity a ¢ of a. Thus geodesics have constant speed, since dif-ferentiation of ||a ¢||2 = a ¢ • a ¢ gives 2a ¢ • a ≤ = 0.

A straight line a(t) = p + tq contained in M is always a geodesic of Msince its acceleration a ≤ = 0 is trivially normal to M. Though they lack anygeometric significance, constant curves are also geodesics, but to avoid clutterthis case is often neglected.

6.9 Example Geodesics of some surfaces in R3.(1) Planes. If a is a geodesic in a plane P orthogonal to u, then a ¢ • u = 0,

hence a ≤ • u = 0. But a ≤ is by definition normal to P, hence collinear withu, so a ≤ = 0. Thus a is a straight line. Since as noted above, every such lineis a geodesic, we conclude that the geodesics of P are the straight lines in P.

(2) Spheres. A great circle in a sphere S Ã R3 is a circle cut from S by aplane P through the center (Fig. 5.32). If a is a constant-speed parametriza-tion of any circle, we know that its acceleration a ≤ points toward the centerof the circle. In the case of a great circle that center is also the center of thesphere S. Thus a ≤ is normal to S, so a is a geodesic of S.

We can find such a geodesic with any given initial velocity vp (the requiredplane P passes through p orthogonal to p ¥ v). Hence by the uniqueness



feature mentioned earlier, this construction yields all the geodesics of S.Explicitly, the geodesics of a sphere are the constant-speed parametrizations ofits great circles (Fig. 5.32).

(3) Cylinders. The geodesics of, say, the circular cylinder M: x2 + y2 = r2

are all curves of the form

To see this, write an arbitrary curve in M as

A vector normal to M must have z coordinate zero. Thus if a is a geodesic,h≤ = 0, so h(t) = ct + d. Since the speed of a geodesic is constant, the speed(r2J¢2 + h¢2)1/2 of a is constant, so J ¢ is constant. Hence J(t) = at + b.

When both constants a and c are nonzero, a is a helix on M. In extremecases, a parametrizes a ruling if a = 0 and a cross-sectional circle if c = 0.◆

A closed geodesic is a geodesic segment a : [a, b] Æ M that is smoothlyclosed (g ¢(b) = g ¢(a)) and hence extendible by periodicity over the whole realline. Thus closed geodesics and periodic geodesics are effectively the samething. In the surfaces above, every geodesic of the sphere is closed, while onthe cylinder only the cross-sectional circles are closed.

6.10 Remark Here is a simple geometric way to find examples of geo-desics. If a unit-speed curve a in M lies in a plane P everywhere orthogonalto M along a, then a is a geodesic of M. Proof. Since a has constant speed,a ≤ is always orthogonal to a ¢, but these two vectors lie in a plane orthogo-nal to M, and a ¢ is always tangent to M. Hence a ≤ must be orthogonal toM, so a is geodesic.

Using this remark we could have found all the geodesics in the precedingexample except the helices in the cylinder. It shows at once that on a surfaceof revolution M, all meridians are geodesics, since they are cut from M byplanes passing through the axis of rotation and hence orthogonal to M.

a J Jt r t r t h t( ) = ( ) ( ) ( )( )cos sin ., ,

a t r at b r at b ct d( ) = +( ) +( ) +( )cos sin ., ,

FIG. 5.32


The essential properties of the three types of curves we have consideredcan be summarized as follows:

Principal curves k(a ¢) = k1 or k2, S(a ¢) collinear a ¢,Asymptotic curves k(a ¢) = 0, S(a ¢) orthogonal to a ¢, a ≤ tangent to MGeodesics a ≤ normal to M

Exercises

1. Prove that a curve a in M is a straight line of R3 if and only if a is bothgeodesic and asymptotic.

2. To which of the three types—principal, asymptotic, geodesic—do the fol-lowing curves belong?

(a) The top circle a of a torus (Fig. 5.33).(b) The outer equator b of a torus.(c) The x axis in M: z = xy.

(Assume constant-speed parametrizations.)

3. (Closed geodesics.) Show:(a) In a surface of revolution, a parallel through a point a(t) on the profilecurve is a (necessarily closed) geodesic if and only if a ¢(t) is parallel to theaxis of revolution.(b) There are at least three closed geodesics on every ellipsoid (Ex. 9 ofSec. 4.2).

4. Let a be an asymptotic curve in M Ã R3 with curvature k > 0.(a) Prove that the binormal B of a is normal to the surface along a, anddeduce that S(T ) = tN.(b) Show that along a the surface has Gaussian curvature K = -t 2.(c) Use (b) to find the Gaussian curvature of the helicoid (Ex. 4.3).

5. Suppose that a curve a lies in two surfaces M and N that make a con-stant angle along a (that is, U • V constant). Show that a is principal in Mif and only if principal in N.

FIG. 5.33


6. If x is a patch in M, prove that a curve a (t) = x(a1(t), a2(t)) is(a) Principal if and only if

(b) Asymptotic if and only if a1¢2 + 2a1¢a2¢ + a2¢2 = 0.

7. Let a be a unit-speed curve in M Ã R3. Instead of the Frenet frame fieldon a, consider the Darboux frame field T, V, U—where T is the unit tangentof a, U is the surface normal restricted to a, and V = U ¥ T (Fig. 5.34).

(a) Show that

where k = S(T ) • T is the normal curvature k(T ) of M in the T direction,and t = S(T ) • V.The new function g is called the geodesic curvature of a.

(b) Deduce that a is

8. If a is a (unit speed) curve in M, show that(a) a is both principal and geodesic if and only if it lies in a plane every-where orthogonal to M along a.(b) a is both principal and asymptotic if and only if it lies in a plane every-where tangent to M along a.

geodesic ,

asymptotic ,

principal

¤ =

¤ =

¤ =

g

k

t

0

0

0.

¢ = +

¢ = - +

¢ = - -

T gV kU

V gT tU

U kT tV

,

,

,

¢ - ¢ ¢ ¢=

a a a a

E F G2

21 2 1

2

0

L M N

.

FIG. 5.34

9. On the monkey saddle M (see Fig. 5.19) find three asymptotic curves andthree principal curves passing through the origin 0. (This is possible onlybecause 0 is a planar umbilic point.)

10. Let a be a regular curve in M Ã R3, and let U be the unit normal of Malong a. Show that a is a principal curve of M if and only if the ruled surfacex(u, v) = a(u) + vU(u) is flat.

11. A ruled surface is noncylindrical if its rulings are always changing direc-tions; thus for any director curve, d ¥ d ¢ π 0. Show that:

(a) a noncylindrical ruled surface has a parametrization

for which ||d || = 1 and s ¢ • d¢ = 0.(b) for this parametrization,

The curve s is called the striction curve, and the function p is the distributionparameter.

(c) Deduce from the behavior of K on each ruling that the route of thestriction curve is independent of parametrization, and hence that the dis-tribution parameter is essentially a function on the set of rulings.

(Hint: For (a), find f such that s = a + fd. For (b), show that s ¢ ¥ d = pd ¢.)

12. In each case below, find the striction curve and distribution parameter,and check the formula for K in (b) of the preceding exercise.

(a) the helicoid in Example 4.3.(b) the tangent surface of a curve (Ex. 13 of Sec. 4).(c) both sets of rulings of the saddle surface in Example 4.3.

(Hint: In the usual ruled parametrization (u, 0, 0) + v(0, 1, u), this last vectormust be replaced by a unit vector in order to apply Ex. 11. The curvatureformula in Ex. 4.3 will then change.)

13. If x(u, v) = a(u) + vd(u) parametrizes a noncylindrical ruled surface,let L(u) be the ruling through a(u). Show that:

(a) If Je is the smallest angle from L(u) to L(u + e), and de is the orthog-onal distance from L(u) to L(u + e), then

Thus the distribution parameter is the rate of turning of L.

lim .e

e

eJÆ= ( )

0

dp u

Kp u

p u vp=

- ( )( ) +( )

¢ ¥ ¢¢ ¢

2

2 2 2 , where =s d d

d d•

•.

x u v u v u,( ) = ( ) + ( )s d



(b) There is a unique point pe of L(u) that is nearest to L(u + e), and

limeÆ0

pe = s(u)

(Fig. 5.35). (This gives another characterization of the striction curve s.)

(Hint: The common perpendicular to L(u) and L(u + e) is in the directionof d(u) ¥ d(u + e) ª ed(u) ¥ d ¢(u).)

14. Let x(u, v) = a(u) + vd(u), with ||d || = 1, parametrize a flat ruled surfaceM. Show that:

(a) If a¢ is always zero, then M is a generalized cone.(b) If d¢ is always zero, then M is a generalized cylinder.(c) If both a ¢ and d ¢ are never zero, then M is the tangent surface of itsstriction curve. (Hint: Parametrize by s + vd as in Ex. 11, giving s unitspeed. Use K = 0 to show that T = s ¢ and d are collinear.)

These are only the extreme cases. For example, a flat piece of paper could bebent cylindrically at one end and conically at the other. Note that of the threetypes, only the cylinder has rulings that are entire straight lines.

15. (Enneper’s minimal surface.) Prove:(a) The mapping x: R2 Æ R3 given by

though not one-to-one, is regular, and hence defines an immersed surface E.(b) x is a principal parametrization of E, that is, the u- and v-parametercurves are principal curves.(c) E is a minimal surface.(d) The asymptotic curves of E are u Æ x(u, ±u).

16. (Continuation by computer graphics.)(a) Plot x(D) Ã E, for D: -3 � u, v � 3. (Note that by the preceding exer-cise the parameter curves are principal.)(b) Show that the Euclidean isometry (x, y, z) Æ (-y, x, -z) carries E to itself.

x u v uu

uv vv

vu u v, , , ,( ) = - + - + -ÊËÁ

ˆ¯

32

32 2 2

3 3

FIG. 5.35


Thus the z < 0 half of E is the mirror image of a 90° rotation of the z > 0half. Further properties of E are developed in Exercise 10 of Section 6.8.

17. A right conoid is a ruled surface whose rulings all pass orthogonallythrough a fixed axis (Fig. 5.36). Taking this axis as the z axis of R3, we findthe parametrization

where the u-parameter curves are the rulings. (This reversal of u and v fromearlier exercises makes it clear that the helicoid is a conoid.)

(a) Find the Gaussian and mean curvature of x.(b) Show that the surface is noncylindrical if J¢ is never zero, and in thiscase, find the striction curve and parameter of distribution.

18. (Computer graphics.)(a) A right conoid has base curve a(v) = (0, 0, cos2v) and director curved (v) = (cosv, sinv, 0). For the resulting ruled parametrization (with u-parameter curves as rulings), plot the portion with -2.5 � u � 2.5, 0 � v� p.(b) The axis of a right conoid is the z axis, and its rulings pass throughevery point of the circle y2 + z2 = r2 in the plane x = c. Verify that

x(u, v) = (uc, urcosv, rsinv)

parametrizes this conoid, and for r = 2, c = 1 plot the portion between the axis and the circle.(c) Same as (b) except that the circle is replaced by the curve y = sinz.Find a ruled parametrization, and for c = 4, plot the region with 0 � z� 4p and rulings running from x = -4 to x = +4.

19. For curves b and d in R3, let x(u, v) = b(u) + vd(u). Find, in simplestterms:

x u v u v u v h v, cos sin( ) = ( ) ( ) ( )( )J J, , ,

FIG. 5.36

(a) A necessary and sufficient condition that x parametrize a (ruled)surface in R3.(b) A formula for the Gaussian curvature K of this surface. (Hint: ShowN = 0.)

5.7 Surfaces of Revolution

The geometry of a surface of revolution is rather simple, yet these surfacesexhibit a wide variety of geometric behavior; thus they offer a good field forexperiment.

We apply the methods of Section 4 to study an arbitrary surface of revolu-tion M, with the usual parametrization, given in Example 2.4 of Chapter 4 by

Here h(u) > 0 is the radius of the parallel at distance g(u) along the axis ofrevolution of M, as shown in Fig. 4.14. This geometric significance for g andh means that our results do not depend on the particular position of M rel-ative to the coordinate axes of R3.

Because g and h are functions of u alone, we can write

and hence

Here E is the square of the speed of the profile curve and hence of all themeridians (u-parameter curves), while G is the square of the speed of the par-allels (v-parameter curves). Next we find, successively,

Taking second derivatives gives

x

x

x

uu

uv

vv

g h v h v

h v h v

h v h v

= ¢¢ ¢¢ ¢¢( )= - ¢ ¢( )= - -( )

, , ,

, , ,

, , ,

cos sin

sin cos

cos sin

0

0

x x

x x

u v

u v

hh hg v hg v

EG F h g h

Ug h

h g v g v

¥ = ¢ - ¢ - ¢( )

¥ = - = ¢ + ¢

=¢ + ¢

¢ - ¢ - ¢( )

, , ,

,

cos sin

, cos , sin .

2 2 2

2 2

1

E g h F G h= ¢ + ¢ = =2 2 20, , .

x

x

u

v

g h v h v

h v h v

= ¢ ¢ ¢( )= -( )

, , ,

, , ,

cos sin

sin cos0

x u v g u h u v h u v, cos sin .( ) = ( ) ( ) ( )( ), ,


5.7 Surfaces of Revolution 253

Hence

Since F = = 0, x is a principal parametrization (Exercise 8 of Section 4),and for the shape operator S derived from U,

This is an analytical proof that the meridians and parallels of a surface ofrevolution are its principal curves. Furthermore, if the corresponding princi-pal curvature functions are denoted by km and kp , instead of k1 and k2, wehave

(1)

Thus the Gaussian curvature of M is

(2)

This formula defines K as a real-valued function on the domain of the profilecurve

By the conventions of Section 4, K(u) is the Gaussian curvature K(x(u, v)) ofM at every point of the parallel through a (u). The same is true for the otherfunctions above. The rotational symmetry of M about its axis of revolutionmeans that its geometry is “constant on parallels”—completely determinedby the profile curve.

In the special case where the profile curve passes at most once over eachpoint of the axis of rotation, we can usually arrange for the function g to besimply g(u) = u (Example 2.8 of Chapter 4). Then the formulas (1) and (2)above reduce to

(3)

kh

hk

h h

Kh

h h

m p=- ¢¢

+ ¢( )=

+ ¢( )

=- ¢¢+ ¢( )

1

1

1

1

2 3 2 2 1 2

2 2

, ,

.

a u g u h u( ) = ( ) ( )( ), , 0 .

K k kg

g h

g h

h g h= =

- ¢¢ ¢¢¢ ¢¢

¢ + ¢( )m p 2 2 2 .

kE

g h

g h

g hk

Gg

h g hm p= =

-¢ ¢¢¢ ¢¢

¢ + ¢( )= =

¢¢ + ¢( )

L N

2 2 3 2 2 2 1 2, .

SE

SGu u v vx x x x( ) = ( ) =

L N, .

L M N, ,=- ¢ ¢¢ + ¢¢ ¢

¢ + ¢= =

¢¢ + ¢

g h g h

g h

g h

g h2 2 2 20 .


7.1 Example Surfaces of revolution.(1) Torus of revolution T. The usual parametrization x in Example 2.5 of

Chapter 4 has

for constants 0 < r < R. Although the axis of revolution is now the z axis,formulas (1) and (2) above remain valid, and we compute

This gives an analytical proof that the Gaussian curvature of the torus is positive on the outer half and negative on the inner half. K has its maximumvalue 1/r(R + r) on the outer equator (u = 0), its minimum value -1/r(R - r)on the inner equator (u = p), and is zero on the top and bottom circles (u = ±p /2).

(2) Catenoid. The curve y = c cosh(x/c) is a catenary; its shape is that ofa chain hanging under the influence of gravity. The surface obtained by rotat-ing this curve around the x axis is called a catenoid (Fig. 5.37). From the for-mulas (3) we find,

E r F G R r u

r R r u u

kr

ku

R r u

Ku

r R r u

= = ( )= = = +( )

= =+

=+( )

2 20

0

1

, , = + cos ,

, , ,

, ,

L M N cos cos

coscos

coscos

.

m p

g u r u h u R r u( ) = ( ) = +sin cos, ,

FIG. 5.37


and hence

Since its mean curvature H is zero, the catenoid is a minimal surface. ItsGaussian curvature interval is -1/c2 � K < 0, with minimum value K = -1/c2

on the central circle (u = 0). ◆

7.2 Theorem If a surface of revolution M is a minimal surface, then Mis contained in either a plane or a catenoid.

Proof. M is parametrized as usual by

with u in a (possibly infinite) interval I.

Case 1. g¢ is identically zero. Then g is constant, so M is part of a planeorthogonal to the axis of revolution.

Case 2. g¢ is never zero. By Exercise 8 in Section 4.2, M has a parame-trization of the form

The formulas for km and kp in (3) above then show that the minimality con-dition is equivalent to

Because u does not appear explicitly in this differential equation, there is astandard elementary way to solve it. We merely record that the solution is

where a π 0 and b are constants. Thus M is part of a catenoid.Case 3. g¢ is zero at some points, nonzero at others. This cannot happen.

For definiteness, suppose that g¢(u) > 0 for u < u0 but g¢(u0) = 0. By Case 2, the profile curve (g(u), h(u), 0) is a catenary for u < u0. The shape of the catenary makes it clear that slope h¢/g¢ cannot approach infinity as u Æ u0. ◆

h u aua

b( ) = +ÊËÁ

ˆ¯cosh ,

hh h¢¢ = + ¢1 2.

y u v u h u v h u v, cos sin( ) = ( ) ( )( ), , .

x u v g u h u v h u v, cos sin( ) = ( ) ( ) ( )( ), , ,

H Kc x c

= =-

( )01

2 4,

cosh.

- = = ( )k kc x cm p

12cosh

,

This result shows that catenoids are the only complete nonplanar surfaces ofrevolution that are minimal. (Completeness, discussed in Chapter 8, impliesthat the surface cannot be part of a larger surface.)

Helicoids and catenoids are called the elementary minimal surfaces. Twoothers are given in the exercises for this chapter (Exercise 5 in Section 5 andExercise 15 in Section 6). Soap-film models of an immense variety of minimalsurfaces can easily be exhibited by the methods given in [dC], where the term“minimal” is explained.

The expression , which appears so frequently in the formulasabove, is just the speed of the profile curve a(u) = (g(u), h(u), 0). Thus wecan radically simplify these formulas by a reparametrization that has unitspeed. The surface of revolution is unchanged; it has merely been given a newparametrization, called canonical.

7.3 Lemma For a canonical parametrization of a surface of revolution,

and the Gaussian curvature is

Proof. Since g¢2 + h¢2 = 1 for a canonical parametrization, these expres-sions for E, F, and G follow immediately from those at the start of thissection. The formula for K in (2) becomes

But this can be simplified. Differentiation of g¢2 + h¢2 = 1 gives g¢ g≤ = -h¢h≤, and when this is substituted above, we get K = -h≤/h. ◆

The effect of using a canonical parametrization is to shift the emphasisfrom measurements in the space outside M (for example, along the axis ofrevolution) to measurement within M. This important idea will be developedmore fully as we proceed.

7.4 Example Canonical parametrization of the catenoid (c = 1).An arc length function for the catenary a(u) = (u, cosh u) is s(u) = sinh u.Hence a unit-speed reparametrization is

b s g s h s s s( ) = ( ) ( )( ) = +( )-, , ,sinh 1 21

Kgh

g h

g hg h g g h

h=

- ¢ ¢ ¢¢¢ ¢¢

=- ¢ ¢¢ + ¢ ¢¢ ¢2

.

Khh

=- ¢¢

.

E F G h= = =1 0 2, , ,

¢ + ¢g h2 2



as indicated in Fig. 5.38. The resulting canonical parametrization of thecatenoid is given by

Hence by the preceding lemma,

This formula for K in terms of x is consistent with the formula

found in Example 6.1 for the parametrization x1. In fact, since s(u) = sinh u,we have

The simple formula for K in Lemma 7.3 suggests a way to construct sur-faces of revolution with prescribed Gaussian curvature. Given a function

K = K(u) on some interval,

first solve the differential equation h≤ + Kh = 0 for h, subject to initial con-ditions h(0) > 0 and |h¢ (0)| < 1. (The first of these conditions is a convenience;the second is a necessity since we must have g¢2 + h¢2 = 1.)

To get a canonical parametrization, we need a function g satisfying theequation g¢2 + h¢2 = 1. Evidently,

will do the job.We conclude that for any interval around 0 on which the initial conditions

h h> ¢ <0 1 and

g u h t dtu

( ) = - ¢ ( )Ú 1 2

0

K s us u u u

( )( ) =-

+ ( )( )=

-+( )

=-1

1

1

1

12 2 2 2 4sinh cosh

.

K uu

( ) =1

4cosh

K sh sh s s

( ) = -¢¢( )( ) =

-+( )

1

1 2 2 .

x s v s s v s v, sinh cos sin .( ) = + +( )-1 2 21 1, ,

FIG. 5.38


both hold, revolving the profile curve (g(u), h(u), 0) around the x axisproduces a surface that has, by Lemma 7.3, Gaussian curvature

A natural use of this scheme is to look for surfaces that have constant cur-vature. Consider first the K positive case.

7.5 Example Surfaces of revolution with constant positive curvature.We apply the procedure to the constant function K = 1/c2. The differential

equation h≤ + h/c2 = 0 has general solution

The constant b represents only a translation of coordinates so we may as wellset b = 0. As usual, nothing is lost by requiring h > 0; hence a > 0. Thus thefunctions

give rise to a surface of revolution Ma with constant Gaussian curvature

As mentioned above, the conditions h > 0 and |h¢| < 1 determine the largestinterval I on which the procedure works. The constant c is fixed, but the con-stant a is at our disposal, and it distinguishes three cases.

Case 1. a = c. Here

(4)

Thus the maximum interval I is and the profile curve (g(u),h(u)) is a semicircle. Revolution about the x axis produces a sphere S of radius c—except for its two points on the axis.

Case 2. 0 < a < c. Here h is positive on the same interval as above and |h¢| < 1 is always true, so g is well defined. The profile curve has the samelength , but it now forms a shallower arch, which rests on the x axis at±a*, where (Fig. 5.39). As a shrinks down from c to 0, one can check that a* increases from c to . The resulting surface ofrevolution, round when a = c, first becomes football-shaped and then growsever thinner, becoming, for a small, a needle of length just less than .

By contrast with Case 1, the intercepts (±a*, 0, 0) cannot be added to Mnow since this surface is actually pointed at each end (Fig. 5.39).

pc

pc /2a g c a* /= ( ) >p 2

pc /2

- < <p pc u c/ / ,2 2

g utc

dt cuc

h u cuc

u

( ) = = ( ) =Ú cos sin , cos .0

K c= 1 2 .

g uac

tc

dt h u auc

u

( ) = -ÊËÁ

ˆ¯ ( ) =Ú 1

2

22

12

sin , cos0

h u auc

b( ) = +ÊËÁ

ˆ¯cos .

K h h= - ¢¢/ .


The differential equation has delicately adjusted the shapeof Ma so that its principal curvatures are no longer equal but still give

Case 3. a > c. Here the maximum interval is shorter than in Case 1. Theformula for g(u) in (4) shows that the endpoints now are ±a*, where a* < c isdetermined by . Thus,

As a increases from a = c, the resulting surface of revolution Ma is at firstsomewhat like the outer half of a torus. But when a is very large, it becomesa huge circular band (Fig. 5.40), whose very short profile curve is sharplycurved (km must be large since ).

A corresponding analysis for constant negative curvature leads to an infinite family of surfaces of revolution with (Exercises 7 and 8).The simplest of these surfaces is

7.6 Example The bugle surface B. The profile curve of B (in the xy plane)is characterized by this geometric condition: It starts at the point (0, c) andmoves so that its tangent line reaches the x axis after running for distanceexactly c. This curve, a tractrix, can be described analytically as

a(u) = (u, h(u)), u > 0,

K c= -1 2/

k a k k cp m pª =1 1 2/ /and

h a a a c a c* cos * .( ) = = -2 2

sin */ /a c c a= < 1

K k kc

= =m p1

2.

¢¢ + =h h c/ 2 0

FIG. 5.40

FIG. 5.39


where h is the solution of the differential equation

such that h(u) Æ c as u Æ 0. The resulting surface of revolution B is calleda bugle surface or tractroid (Fig. 5.41). Using the differential equation above,we deduce from the earlier formulas (3) that the principal curvatures of B are

Thus the bugle surface has constant negative curvature

This surface cannot be extended across its rim—not part of B—to form alarger surface in R3 since km(u) Æ • as u Æ 0. ◆

When this surface was first discovered, it seemed to be the analogue, for Ka negative constant, of the sphere; it was thus called a pseudosphere. However,as we shall see later on, the true analogue of the sphere is quite a differentsurface and cannot be found in R3.

Exercises

1. Find the Gaussian curvature of the surface obtained by revolving thecurve around the x axis. Sketch this surface and indicate the regionswhere K > 0 and K < 0.

y e x= - 2 2

Kc

= -1

2.

khc

kchm p=

- ¢=

¢,

1.

¢ =-

-h

h

c h2 2

FIG. 5.41


2. (a) Show that when y = f(x) is revolved around the x axis, the Gaussiancurvature K(x) has the same sign (-, 0, +) as -f ≤(x) for all x.(b) Deduce that for a surface of revolution with arbitrary axis, the Gauss-ian curvature K is positive on parallels through convex intervals on theprofile curve (where the curve bulges away from the axis) and negative onparallels through concave intervals (where the curve sags toward the axis).

3. Prove that a flat surface of revolution is part of a plane, cone, or cylinder.

4. (Computer.)(a) Write computer commands that, given a profile curve u Æ (g(u), h(u)),(i) plot the resulting surface of revolution for a � u � b, and (ii) return itsGaussian curvature K(u).(b) Test (a) on the torus and catenoid in Example 7.1.

5. If is the usual polar coordinate function on the xy plane,and f is a differentiable function, show the M: z = f(r) is a surface of revo-lution and that its Gaussian curvature K is given by

6. Find the Gaussian curvature of the surface M: . Sketch thissurface, indicating the regions where K > 0 and K < 0.

7. (Surfaces of revolution with negative curvature ) As in the cor-responding positive case, there is a family of such surfaces, separated intotwo subfamilies by a special surface. Essentially all these surfaces are given,using canonical parametrization, by solutions of as follows:

(a) If 0 < a < c, let Ma be the surface given by h(u) = a sinh , u > 0.Show that its profile curve (g(u), h(u)) leaves the origin with slope

and rises to a maximum height of .(b) If a = c, let B

–be the surface given by h(u) = ceu/c, u < 0. Show that

its mirror image B, given by h(u) = ce-u/c, u > 0, is the bugle surface inExample 7.6.

c a2 2-a c a2 2-

u c/( )¢¢ - =h h c/ 2 0

K c= -1 2/

z e r= - 2 2

K rf r f r

r f r( ) =

¢( ) ¢¢( )+ ¢( )( )1 2 2 .

r x y= +2 2

FIG. 5.42

(c) If b > c, let Mb be the surface given by . Show thatas |u| increases from 0, its profile curve rises symmetrically from height b

to

height .

Sample profile curves of all three types are shown in Fig. 5.42, where Ma andMb have been translated along the axis of revolution. Explicit formulas forthe profile curves in (a) and (b) involve elliptic integrals (see [G]).

8. (a) Taking c = 1 for simplicity, show that the tractrix has a parame-trization (g, h) with

(b) (Computer graphics.) Plot a view of the resulting bugle surface similarto that in Fig. 5.41.

9. In a twisted surface of revolution, as points rotate around the axis theyalso move evenly in the axis direction. Explicitly, if the original surface hasa usual parametrization in terms of functions g(u) and h(u), then the twistedsurface has parametrization

where p is a constant.(a) Find a parametrization of the twisted bugle surface D (Dini’s surface) with data as in the preceding exercise and .(b) (Computer.) Plot the surface D in (a) for 0.01 � u � 2 and 0 � v � 6p.(Impose smoothness and view the surface from a point with x < 0.)(c) Show that D has constant negative curvature.

5.8 Summary

The shape operator S of a surface M in R3 measures the rate of change of aunit normal U in any direction on M and thus describes the way the shapeof M is changing in that direction. If we imagine U as the “first derivative”of M, then S is the “second derivative.” But the shape operator is an alge-braic object consisting of linear operators on the tangent planes of M. Andit is by an algebraic analysis of S that we have been led to the main geomet-ric invariants of a surface in R3: its principal curvatures and directions, andits Gaussian and mean curvatures.

p = 1 5

x u v g u pv h u v h u v, cos sin .( ) = ( ) ( ) ( )( )+ , ,

g u e h u e eu u u( ) = ( ) = - - + -- - -, h1 12 2arctan .

c b2 2+

h u b u c( ) = ( )cosh /


▼

▲Chapter 6

Geometry of Surfaces in R3

263

Now that we know how to measure the shape of a surface M in R3, the nextstep is to see how the shape of M is related to its other properties. Near eachpoint of M, the Gaussian curvature has a strong influence on shape (Remark3.3 of Chapter 5), but we are now interested in the situation in the large—over the whole extent of M. For example, what can be said about the shapeof M if it is compact, or flat, or both?

In the early 1800s Gauss raised a question that led to a new and deeperunderstanding of what geometry is: How much of the geometry of a surfacein R3 is independent of its shape? At first glance this seems a strange ques-tion—what can we possibly say about a sphere, for example, if we ignore thefact that it is round? To get some grip on Gauss’s question, let us imaginethat the surface M Ã R3 has inhabitants who are unaware of the space outsidetheir surface, and thus have no conception of its shape in R3. Nevertheless,they will still be able to measure the distance from place to place in M andfind the area of regions in M. We shall see that, in fact, they can constructan intrinsic geometry for M that is richer and no less interesting than thefamiliar Euclidean geometry of the plane R2.

6.1 The Fundamental Equations

To study the geometry of a surface M in R3 we shall apply the Cartanmethods outlined in Chapter 2. As with the Frenet theory of a curve in R3,this requires that we put frames on M and examine their rates of change alongM. Formally, a Euclidean frame field on M Ã R3 consists of three Euclideanvector fields (Definition 3.7, Chapter 4) that are orthonormal at each point.Such a frame field can be fitted to its surface as follows.

1.1 Definition An adapted frame field E1, E2, E3 on a region O in M Ã R3

is a Euclidean frame field such that E3 is always normal to M (hence E1 andE2 are tangent to M) (Fig. 6.1).

Thus the normal vector field denoted by U in the preceding chapter nowbecomes E3. For brevity we often refer to an adapted frame field “on M,” butthe actual domain of definition is in general only some region in M, since anadapted frame field need not exist on all of M.

1.2 Lemma There is an adapted frame field on a region O in M Ã R3 ifand only if O is orientable and there exists a nonvanishing tangent vectorfield on O.

Proof. This condition is certainly necessary, since E3 orients O, and E1

and E2 are unit tangent vector fields. To show that it is sufficient, let O beoriented by a unit normal vector field U, and let V be a tangent vector fieldthat does not vanish on O. But then it is easy to see that

is an adapted frame field on O. ◆

1.3 Example Adapted frame fields.(1) Cylinder M: x2 + y2 = r2. The gradient of g = x2 + y2 leads to the unit

normal vector field Obviously the unit vector field U3 istangent to M at each point. Setting E2 = U3 ¥ E3, we then get the adaptedframe field

E xU yU r3 1 2= +( ) .

EVV

E U E E U1 2 1 3= = ¥ =, ,

264 6. Geometry of Surfaces in R3

FIG. 6.1

on the whole cylinder M (Fig. 6.2).

(2) Sphere S: x2 + y2 + z2 = r2. The outward unit normal

is defined on all of S, but as we shall see in Chapter 7, every tangent vectorfield on S must vanish somewhere. For example, the “due east” vector fieldV = -yU1 + xU2 is zero at the the north and south poles (0, 0, ±r). Thus theadapted frame field

(Fig. 6.3) is defined on the region O in S gotten by deleting the north andsouth poles.

EVV

E E E

ExU yU zU

r

1

2 3 1

31 2 3

=

= ¥

=+ +

,

,

ExU yU zU

r31 2 3=

+ +

E U

EyU xU

r

ExU yU

r

1 3

21 2

31 2

=

=- +

=+

,

,

6.1 The Fundamental Equations 265

FIG. 6.2

FIG. 6.3

Lemma 1.2 implies in particular that there is an adapted frame field on theimage x(D) of any patch in M; thus such fields exist locally on any surface in R3.

Now we shall bring the connection equations (Theorem 7.2 of Chapter 2)to bear on the study of a surface M in R3. Let E1, E2, E3 be an adapted framefield on M. By moving each frame E1(p), E2(p), E3(p) over a short interval onthe normal line at each point p, we can extend the given frame field to onedefined on an open set in R3. Thus the connection equations

are available for use. We shall apply them only to vectors v tangent to M. Inparticular, the connection forms wij, become 1-forms on M in the sense ofSection 7 of Chapter 2. Thus we have

1.4 Theorem If E1, E2, E3 is an adapted frame field on M Ã R3, and v istangent to M at p, then

The usual interpretation of the connection forms may be read from theseequations, and it bears repetition: wij(v) is the initial rate at which Ei rotatestoward Ej as p moves in the v direction. Since E3 is a unit normal vector fieldon M, the shape operator of M can be described by connection forms.

1.5 Corollary Let S be the shape operator gotten from E3, where E1, E2,E3 is an adapted frame field on M Ã R3. Then for each tangent vector v toM at p,

Proof. By definition, S(v) = -—vE3. The connection equation for i = 3 thengives the result, since the connection form w = (wij) is skew-symmetric:wij = -wji. ◆

In addition to its connection forms, the adapted frame field E1, E2, E3 alsohas dual 1-forms q1, q2, q3 (Definition 8.1 of Chapter 2) that give the co-ordinates qi(v) = v • Ei(p) of any tangent vector vp with respect to the frameE1(p), E2(p), E3(p). As with the connection forms, the dual forms will beapplied only to vectors tangent to M, so they become forms on M. Thisrestriction is fatal to q3, for if v is tangent to M, it is orthogonal to E3, soq3(v) = v • E3(p) = 0. Thus q3 is identically zero on M.

S E Ev v p v p( ) = ( ) ( ) + ( ) ( )w w13 1 23 2 .

— = ( ) ( ) £ £( )=

Âv i ij jj

E E iw v p1

3

1 3 .

— = ( ) ( )Âv i ij jE Ew v p


Because of the skew-symmetry of the connection form, we are left withessentially only five 1-forms:

q1, q2 provide a dual description of the tangent vector fields E1, E2;w12 gives the rate of rotation of E1, E2;w13, w23 describe the shape operator derived from E3.

1.6 Example The sphere. Consider the adapted frame field E1, E2, E3

defined on the (doubly punctured) sphere S in Example 1.3. By extending this frame field to an open set of R3 we get the spherical frame field given in Example 6.2 of Chapter 2, provided the indices of the latter are shifted by1 Æ 3, 2 Æ 1, 3 Æ 2. Thus, in terms of the spherical coordinate functions,Example 8.4 of Chapter 2 gives

Because all forms (including functions) are now restricted to the surface S,the spherical coordinate function r has become a constant: the radius r ofthe sphere.

In general, the forms associated with an adapted frame field obey the fol-lowing remarkable set of equations.

1.7 Theorem If E1, E2, E3 is an adapted frame field on M Ã R3, then itsdual forms and connection forms on M satisfy:

(1)

(2)

(3)

(4)

Proof. We merely apply the Cartan structural equations in Theorem 8.3of Chapter 2. The first structural equation,

d i ijj

jq w q= ŸÂ

dd

First structural equations

Symmetry equation

d Gauss equation

dd

Codazzi equations

q w qq w q

w q w q

w w w

w w ww w w

1 12 2

2 21 1

31 1 32 2

12 13 32

13 12 23

23 21 13

0

= Ÿ= Ÿ

ÏÌÓŸ + Ÿ =

= Ÿ

= Ÿ= Ÿ

ÏÌÓ

q j J w j Jq j w j J

w j

1 12

13

23

= == = -

= -

r d d

r d d

d

cos sin

cos

, ,

, ,

.2

6.1 The Fundamental Equations 267

yields (1) and (2) above. In fact, for i = 1, 2, we get (1), since q3 = 0 on thesurface M. But q3 = 0 implies dq3 = 0, so for i = 3 we get (2).

Then the second structural equation yields the Gauss (3) and Codazzi(4) equations. ◆

Because connection forms are skew-symmetric and a wedge product of 1-forms satisfies f Ÿ y = -y Ÿ f, the fundamental equations above can berewritten in a variety of equivalent ways. However we shall stick to the indexpattern used above, which, on the whole, seems the easiest to remember.

We emphasize that the forms introduced in this section describe not thesurface M directly, but only the particular adapted frame field E1, E2, E3 fromwhich they are derived: A different choice of the frame field will produce dif-ferent forms. Nevertheless, the six fundamental equations in Theorem 1.7contain a tremendous amount of information about the surface M Ã R3, andwe shall call on each in turn as we come to a geometric situation that itgoverns. For example, since w13 and w23 describe the shape operator of M,the Codazzi equations (4) express the rate at which the shape of M is chang-ing from one point to another.

The first of the following exercises shows how the Cartan approach auto-matically singles out the three types of curves considered in Chapter 5,Section 6.

Exercises

1. Let a be a unit-speed curve in M Ã R3. If E1, E2, E3 is an adapted framefield such that E1 restricted to a is its unit tangent T, show that

(a) a is a geodesic of M if and only if w12(T) = 0.(b) If E3 = E1 ¥ E2, then

where g, k, and t are the functions defined in Exercise 7 of Section 5.6.(Hint: If T = E1 along a, then Ei¢ = —E1Ei along a.)

2. (Sphere.) For the frame field in Example 1.6:(a) Verify the fundamental equations (Thm. 1.7).(b) Deduce from the formulas for q1 and q2 that

(c) Use Corollary 1.5 to find the shape operator S of the sphere.

Er

E

E Er

1 1

2 2

10

01

Jj

j

J j

[ ] = [ ] =

[ ] = [ ] =

cos,

.

,

,

g T k T t T= ( ) = ( ) = ( )w w w12 13 23, , ,


3. Give a new proof that shape operators are symmetric by using the sym-metry equation (Thm. 1.7).

6.2 Form Computations

From now on, our study of the geometry of surfaces will be carried on mostlyin terms of differential forms, so the reader may wish to look back over theirgeneral properties in Sections 4 and 5 of Chapter 4. Increasingly, we shalltend to compare M with the Euclidean plane R2. Thus, if E1, E2, E3 is anadapted frame field on M Ã R3, we say that E1, E2 constitutes a tangent framefield on M. Any tangent vector field V on M may be expressed in terms ofE1 and E2 by the orthonormal expansion

To show that two forms are equal, we do not have to check their values onall tangent vectors, but only on the “basis” vector fields E1, E2. (See theremarks preceding Example 4.7 of Chapter 4). Explicitly: 1-forms f and y areequal if and only if

2-forms m and v are equal if and only if

The dual forms q1, q2 are, as we have emphasized, merely another descrip-tion of the tangent frame field E1, E2; they are completely characterized bythe equations

These forms provided a “basis” for the forms on M (or, strictly speaking, onthe region of definition of E1,E2).

2.1 Lemma (The Basis Formulas) Let q1, q2 be the dual 1-forms of E1,E2 on M. If f is a 1-form and m a 2-form, then

(1) f = f(E1)q1 + f(E2)q2,(2) m = m(E1, E2)q1 Ÿ q2.

Proof. Apply the equality criteria above, observing for (2) that by defin-ition of the wedge product,

q di j ijE i j( ) = £ £( )1 2, .

m E E v E E1 2 1 2, ,( ) = ( ).

f y f yE E E E1 1 2 2( ) = ( ) ( ) = ( )and ;

V V E E V E E= ( ) + ( )• • .1 1 2 2

6.2 Form Computations 269

◆

Assuming throughout that the forms q1, q2, w12, w13, w23 derive as in Section1 from an adapted frame field E1, E2, E3 on a region in M, let us see whatsome of the concepts introduced in Chapter 5 look like when expressed interms of forms. We begin with the analogue of Lemma 3.4 of Chapter 5.

2.2 Lemma (1) w13 Ÿ w23 = Kq1 Ÿ q2

(2) w13 Ÿ q2 + q1 Ÿ w23 = 2Hq1 Ÿ q2.

Proof. To apply the definitions K = det S, 2H = trace S, we shall find thematrix of S with respect to E1 and E2. As in Corollary 1.5, the connectionequations give

Thus the matrix of S is

Now, using the second formula in Lemma 2.1, what we must show is that (w13 Ÿ w23) (E1, E2) = K and (w13 Ÿ q2 + q1 Ÿ w23) (E1, E2) = 2H. But

and a similar computation gives the trace formula. ◆

Comparing the first formula above with the Gauss equation (3) in Theorem1.7, we get

2.3 Corollary dw12 = -Kq1 Ÿ q2.

We shall call this the second structural equation,† and derive from it a newinterpretation of Gaussian curvature: w12 measures the rate of rotation of

w w w w w w13 23 1 2 13 1 23 2 13 2 23 1Ÿ( )( ) = ( ) ( ) - ( ) ( )= = =

E E E E E E

S S K

,

determinant of matrix of ,det

w ww w

13 1 13 1

23 2 23 2

E E

E E

( ) ( )( ) ( )

ÊËÁ

ˆ¯.

S E E E E E E

S E E E E E E

E

E

1 3 31 1 1 32 1 2

2 3 31 2 1 32 2 2

1

2

( ) = -— = - ( ) - ( )( ) = -— = - ( ) - ( )

w w

w w

,

.

q q q q q q1 2 1 2 1 1 2 2 1 2 2 1

1 1 0 0 1

Ÿ( )( ) = ( ) ( ) - ( ) ( )= ◊ - ◊ =

E E E E E E,

.


† This equation will be shown to be the analogue for M of the second structural equation for R3

(Theorem 8.3 of Chapter 2).

tangent frame field E1, E2—and since K determined the exterior derivativedw12, it becomes a kind of “second derivative” of E1, E2.

For example, on a sphere of radius r, the formulas in Example 1.6 give

But

Thus the second structural equation gives the expected result,As we have emphasized, a frame field on a surface is most efficient when

it is derived in some natural way from the geometry of that surface, as withthe Frenet frame field in the analogous case of a curve. Here is an importantexample.

2.4 Definition A principal frame field on M Ã R3 is an adapted framefield E1, E2, E3 such that at each point E1 and E2 are principal vectors of M.

So long as its domain of definition contains no umbilics, a principal framefield is uniquely determined—except for changes of sign—by the two princi-pal directions at each point.

Occasionally it may be possible to get a principal frame field on an entiresurface. For example, on a surface of revolution, we can take E1 tangent tomeridians, E2 tangent to parallels. In general, however, about the best we cando is as follows.

2.5 Lemma If p is a nonumbilic point of M Ã R3, then there exists a prin-cipal frame field on some neighborhood of p in M.

Proof. Let F1, F2, F3 be an arbitrary adapted frame field on a neighbor-hood N of p. Since p is not umbilic, we can assume (by rotating F1, F2 ifnecessary) that F1(p) and F2(p) are not principal vectors at p. By hypothe-sis k1(p) π k2(p); hence by continuity, k1 and k2 remain distinct near p. Ona small enough neighborhood N of p, all these conditions are thus in force.

Let Sij be the matrix of S with respect to F1, F2. It is now just a stan-dard problem in linear algebra to compute—simultaneously at all points ofN—eigenvectors of S, that is, principal vectors of M. In fact, at each pointthe tangent vector fields.

V S F k S F

V k S F S F

1 12 1 1 11 2

2 22 1 12 2

= + -( )= -( ) +

,

2

K r= 1 2/ .

d d d d d d dw j J j J j j J12 = ( ) = ( ) Ÿ =sin sin cos .

q q j J j j j J1 22 2Ÿ = = -r d d r d dcos cos .

6.2 Form Computations 271

give eigenvectors of S. (This can be checked by a direct computation ifone does not care to appeal to linear algebra.) Furthermore, the function S12 = S(F1) • F2 is never zero on our selected neighborhood N, so ||V1|| and||V2|| are never zero. Thus the vector fields

consist only of principal vectors, so E1, E2, E3 = F3 is a principal frame fieldon N. ◆

If E1, E2, E3 is a principal frame field on M, then the vector fields E1 andE2 consist of eigenvectors of the shape operator derived from E3. Thus wecan label the principal curvature functions so that S(E1) = k1E1 and S(E2) = k2E2. Comparison with Corollary 1.5 then yields

Thus the basis formula (1) in Lemma 2.1 gives

(*)

This leads to an interesting version of the Codazzi equations.

2.6 Theorem If E1, E2, E3 is a principal frame field on M Ã R3, then

Proof. The Codazzi equations (Theorem 1.7) read

The proof is now an exercise in the calculus of forms as discussed inChapter 4, Section 4. Substituting from (*) above in the first of these equa-tions, we get

hence

If we substitute the structural equation dq1 = w12 Ÿ q2, this becomes

dk k k1 1 2 1 12 2Ÿ = -( ) Ÿq w q .

dk k d k1 1 1 1 2 12 2Ÿ + = Ÿq q w q .

d k k1 1 12 2 2q w q( ) = Ÿ ;

d dw w w w w w13 12 23 23 21 13= Ÿ = Ÿ,

E k k k E

E k k k E

1 2 1 2 12 2

2 1 1 2 12 1

[ ] = -( ) ( )[ ] = -( ) ( )

w

w

,

.

w q w q13 1 1 23 2 2= =k k, .

w w

w w

13 1 1 2

23 1 2 2

0

0

E k E

E E k

( ) = ( ) =

( ) = ( ) =

, ,

,

13

23 .

EVV

EVV1

1

12

2

2

= =,


Now apply these 2-forms to the pair of vector fields E1, E2 to obtain

hence

The other required equation derives in the same way from the Codazziequation dw23 = w21 Ÿ w13. ◆

Note that for a principal frame field, w12(v) tells how the principal direc-tions are changing in the v direction.

Exercises

1. (Cylinder.) Restricting the cylindrical frame field (Example 6.2(1) in Ch.2) to the cylinder x2 + y2 = r2 and reversing indices 1 and 3 gives the adaptedframe field in Example 1.3(1). In Chapter 2 we found that the only nonzeroconnection forms of this frame field were w23 = -dJ = -w32 (new indices).

(a) Express the dual forms q1, q2 of this frame field in terms of cylindricalcoordinates r, J, z. (Hint: See end of Sec. 8, Ch. 2.)(b) Verify the fundamental equations (Thm. 1.7) in this case.(c) Express the shape operator of M in terms of cylindrical coordinates,showing that the frame field in (a) is principal.(d) Use Lemma 2.2 to find K and H.

2. (a) If E1, E2 is a tangent frame field on M with connection form w12,show that

(Hint: Write w12 = f1q1 + f2q2, where fi = w12(Ei), and use Cor. 2.3.)(b) Check this formula on the sphere in Example 1.6.

6.3 Some Global Theorems

We have claimed all along that the shape operator S is the analogue for asurface in R3 of the curvature and torsion of a curve in R3. Simple hypothe-ses on k and t singled out some important types of curves. Let us see now

K E E E E E E= ( )[ ] - ( )[ ] - ( ) - ( )2 12 1 1 12 2 12 12

12 22w w w w .

E k dk E k k E2 1 1 2 1 2 12 1[ ] = ( ) = -( ) ( )w .

0 01 2 2 1 12 1- ( ) = -( ) ( ) -dk E k k Ew ;

6.3 Some Global Theorems 273

what can be done with S in the case of surfaces. (Throughout this section,surfaces are assumed to be connected.)

3.1 Theorem If its shape operator is identically zero, then M is part ofa plane in R3.

Proof. By the definition of shape operator, S = 0 means that any unitnormal vector field E3 on M is Euclidean parallel, and hence can be iden-tified with a point of R3 (see Fig. 6.4).

Choose any point p in M. We will show that M lies in the plane throughp orthogonal to E3. If q is an arbitrary point of M, then since M is con-nected, there is a curve a in M from a(0) = p to a(1) = q. Consider thefunction

But then

hence f is identically zero. In particular,

so every point q of M is in the required plane (Fig. 6.4). ◆

We saw in Chapter 5, Section 3 that requiring a single point p of M Ã R3

to be planar (k1 = k2 = 0, or equivalently S = 0) produces no significant effecton the shape of M near p. But the result above shows that if every point isplanar, then M is, in fact, part of a plane.

f E1 03( ) = -( ) =q p • ,

dfdt

E f= ¢ = ( ) =a • ;3 0 0 0and

f t t E( ) = ( ) -( )a p • .3


FIG. 6.4

Perhaps the next simplest hypothesis on a surface M in R3 is that at eachpoint p, the shape operator is merely scalar multiplication by some number—which a priori may depend on p. This means that M is all-umbilic, that is,consists entirely of umbilic points.

3.2 Lemma If M is an all-umbilic surface in R3, then M has constantGaussian curvature K � 0.

Proof. Let E1, E2, E3 be an adapted frame field on some region O in M.Since M is all-umbilic, the principal curvature functions on O are equal,k1 = k2 = k, and furthermore, E1, E2, E3 is actually a principal frame field(since every direction on M is principal). Thus we can apply Theorem 2.6to conclude that E1[k] = E2[k] = 0. Alternatively, we may write

so by Lemma 2.1, dk = 0 on O. But K = k1k2 = k2, so dK = 2k dk = 0 onO. Since every point of M is in such a region O, we conclude that dK = 0on all of M. It follows that K is constant (Exercise 4 of Section 4.7). ◆

3.3 Theorem If M Ã R3 is all-umbilic and K > 0, then M is part of asphere in R3 of radius .

Proof. Pick at random a point p in M and a unit normal vector E3(p) toM at p. We shall prove that the point

is equidistant from every point of M. (Here k(p) = k1(p) = k2(p) is the prin-cipal curvature corresponding to E3(p).)

Now let q be any point of M, and let a be a curve segment in M froma(0) = p to a(1) = q. Extend E3(p) to a unit normal vector field E3 on a,as shown in Fig. 6.5, and consider the curve

Here we understand that the principal curvature function k derives fromE3; thus k is continuous. But K = k2, and by the preceding lemma, K isconstant, so k is constant. Thus

¢ = ¢ + ¢g a1

3kE .

g a= +1

33

kE in R .

c pp

p= + ( ) ( )13k

E

1/ K

dk E dk E1 2 0( ) = ( ) = ,


But

since by the all-umbilic hypothesis, S is scalar multiplication by k. Thus

so the curve g must be constant. In particular,

so for every point q of M. Since K = k1k2 = k2, we have shown that M is contained in the sphere of center c and radius

Using all three of the preceding results, we conclude that a surface M inR3 is all-umbilic if and only if M is part of a plane or a sphere.

3.4 Corollary A compact all-umbilic surface M in R3 is an entire sphere.

Proof. By the preceding remark, we deduce from Exercise 12 of Section4.7 that M must be an entire plane or sphere. The former is impossible,since M is compact, but planes are not. ◆

We now find a useful geometric consequence of a topological assumptionabout surfaces in R3.

1/ .Kd kc q, /( ) = 1

c q q= ( ) = ( ) = + ( )g g0 11

3kE ,

¢ = ¢ + - ¢( ) =g a a1

0k

k ,

¢ = - ¢( ) = - ¢E S k3 a a ,


FIG. 6.5

◆

3.5 Theorem On every compact surface M in R3 there is a point at whichthe Gaussian curvature K is strictly positive.

Proof. Consider the real-valued function f on M such that f(p) = || p ||2.Thus in terms of the natural coordinates of R3, . Now, f is dif-ferentiable, hence continuous, and M is compact. Thus by Lemma 7.3 ofChapter 4, f takes on its maximum at some point m of M. Since f mea-sures the square of the distance to the origin, m is simply a point of M atmaximum distance r = || m || > 0 from the origin. Intuitively, it is clear thatM is tangent at p to the sphere S of radius r—and that M lies inside S,and hence is more curved than S (Fig. 6.6). Thus we would expect that

. Let us prove this inequality.Given any unit tangent vector u to M at the maximum point m, pick a

unit-speed curve a in M such that a(0) = m, a ¢(0) = u. It follows from thederivation of m that the composite function f(a) also has a maximum at t = 0. Thus

(1)

But Evaluating at t = 0, we find

Since u was any unit tangent vector to M at m, this means that m (consid-ered as a vector) is normal to M at m.

Differentiating again, we get

d fdt

2

22 2

aa a a a

( )= ¢ ¢ + ¢¢• • .

0 0 2 0 0 2=( ) ( ) = ( ) ¢( ) =

d fdt

aa a• • .m u

fd f

dta a a

aa a( ) =

( )= ¢• , • .so 2

ddt

fddt

fa a( )( ) = ( )( ) £0 0 0 02

2, .

K rm( ) ≥ >1 02/

f xi= Â 2


FIG. 6.6

By (1), at t = 0 this yields

(2)

The discussion above shows that may be considered as a unit normalvector to M at m, as shown in Fig. 6.7. Thus is precisely the normal curvature k(u) of M in the u direction, and it follows from (2)that In particular, both principal curvatures satisfy thisinequality, so

◆

Thus there are no compact surfaces in R3 with K � 0.Maintaining the hypothesis of compactness, we consider the effect of

requiring that Gaussian curvature be constant. Theorem 3.5 shows that theonly possibility is K > 0. Spheres are obvious examples of compact surfacesin R3 with constant positive Gaussian curvature. It is one of the most remark-able facts of surface theory that they are the only such surfaces. To prove thiswe need a nontrivial preliminary result.

3.6 Lemma (Hilbert) Let m be a point of M Ã R3 such that(1) k1 has a local maximum at m;(2) k2 has a local minimum at m;(3) k1(m) > k2(m).

Then K(m) � 0.

Kr

m( ) ≥ >1

02

.

k ru( ) -� 1/ .

m / •r ¢¢am /r

0 0

1 0

≥ + ¢¢( )= + ¢¢( )

u u m

m

• •

• .

a

a


FIG. 6.7

For example, it is easy to see that these hypotheses hold at any point onthe inner equator of a torus or on the minimal circle (x = 0) of the catenoid.And K is, in fact, negative in both these examples.

To convert hypotheses (1) and (2) into usable form in the proof that follows,we recall some facts about maxima and minima. If f is a (differentiable) func-tion on a surface M and V is a tangent vector field, then the first derivativeV [ f ] is again a function on M. Thus we can apply V again to obtain thesecond derivative V [V [ f ]] = VV [ f ]. A straightforward computation showsthat if f has a local maximum at a point m, then the analogues of the usualconditions in elementary calculus hold, namely,

For a local minimum, of course, the inequality is reversed.

Proof. Since k1(m) > k2(m), m is not umbilic; hence by Lemma 2.5 thereexists a principal frame field E1, E2, E3 on a neighborhood of m in M. Bythe remark above, the hypotheses of minimality and maximality at m implyin particular

(1)

and

(2)

Now we use the Codazzi equations (Theorem 2.6). From (1) it follows that

since k1 - k2 π 0 at m. Thus by Exercise 2(a) of Section 2,

(3)

Applying E1 to the first Codazzi equation in Theorem 2.6 yields

But at the special point m, we have w12(E2) = 0 and k1 - k2 > 0; hencefrom (2) we deduce

(4)

A similar argument starting from the second Codazzi equation gives

(5)E E2 12 1 0w ( )[ ] £ at m.

E E1 12 2 0w ( )[ ] ≥ at m.

E E k E k E k E k k E E1 1 2 1 1 1 2 12 2 1 2 1 12 2[ ] = [ ] - [ ]( ) ( ) + -( ) ( )[ ]w w .

K E E E E= ( )[ ] - ( )[ ]2 12 1 1 12 2w w at m.

w w12 1 12 2 0E E( ) = ( ) = at m

E E k E E k1 1 2 2 2 10 0[ ] ≥ [ ] £and at m.

E k E k1 2 2 1 0[ ] = [ ] = at m

V f VV f[ ] = [ ] £0 0, at m.


Using (4) and (5) in the expression (3) for the Gaussian curvature at m, weconclude that K(m) � 0. ◆

3.7 Theorem (Liebmann) If M is a compact surface in R3 with constantGaussian curvature K, then M is a sphere of radius . (Theorem 3.5implies K is positive.)

Proof. Since M Ã R3 is compact, Theorem 7.10 of Chapter 4 shows it isorientable, so a unit normal exists with shape operators S that are smoothlydefined on the entire surface. Thus continuous principal curvature func-tions k1 � k2 � 0 are also globally well-defined. By Lemma 7.3 of Chapter4, k1 has a maximum at some point p of M. Since K = k1k2 is constant, k2

has a minimum there. Now it cannot be true that k1(p) > k2(p), for thenHilbert’s lemma would give K � 0. Thus k1(p) = k2(p). By the choice of p,it follows that M is all umbilic; hence it is a standard sphere of radius

◆

Liebmann’s theorem is false if the compactness hypothesis is omitted, forwe saw in Chapter 5, Section 7 that there are many nonspherical surfaces withconstant positive curvature.

Exercises

1. If M is a flat minimal surface, prove that M is part of a plane.

2. Flat surfaces in R3 can be bent only along straight lines (see Fig. 6.9). Ifk1 = 0 but k2 is never zero, show that the principal curves of k1 are line seg-ments in R3. (Hint: With {Ei} principal and a≤ = —E1E1, use Thm. 1.4.)

3. Let M Ã R3 be a compact surface with K > 0. If M has constant meancurvature H, show that M is a sphere of radius .

4. Prove that in a region free of umbilics there are two principal curvesthrough each point, these crossing orthogonally. (Hint: Use Ex. 14 of Sec.4.8.)

5. If the principal curvatures of a surface M Ã R3 are constant, show thatM is part of either a plane, a sphere, or a circular cylinder. (In the case k1 π k2, assume there is a principal frame field on all of M.)

1/ H

r = 1/ .K

1/ K


6.4 Isometries and Local Isometries

We remarked earlier that the inhabitants of a surface M in R3, unaware ofthe space outside their surface, could nevertheless determine the distance inM between any two points of M—just as distance on the surface of the earthcan be determined by its inhabitants. The mathematical formulation is asfollows.

4.1 Definition If p and q are points of M Ã R3, consider the collectionof all curve segments a in M from p to q. The intrinsic distance r(p, q) fromp to q in M is the greatest lower bound of the lengths L(a) of these curvesegments.

There need not be a curve a whose length is exactly r(p, q) (see Exercise 3).The intrinsic distance r(p, q) will generally be greater than the straightlineEuclidean distance d(p, q), since the curves a are obliged to stay in M(Fig. 6.8).

On the surface of the earth (sphere of radius about 4000 miles) it is, ofcourse, intrinsic distance that is of practical interest. One says, for example,that it is 12,500 miles from the north pole to the south pole, though theEuclidean distance through the center of the earth is only 8000 miles.

We saw in Chapter 3 how Euclidean geometry is based on the notion ofisometry, a distance-preserving mapping. For surfaces in M we shall prove thedistance-preserving property and use its infinitesimal form (Corollary 2.2 ofChapter 3) as the definition.

6.4 Isometries and Local Isometries 281

FIG. 6.8

4.2 Definition An isometry F: M Æ of surfaces in R3 is a one-to-onemapping of M onto that preserves dot products of tangent vectors. Explic-itly, if F* is the derivative map of F, then

for any pair of tangent vectors v, w to M.

If F* preserves dot products, then it also preserves lengths of tangentvectors. It follows that an isometry is a regular mapping (Chapter 4, Section5), for if F*(v) = 0, then

hence v = 0. Thus by the remarks following Theorem 5.4 of Chapter 4, anisometry F: M Æ is in particular a diffeomorphism, that is, has an inversemapping F-1: Æ M. Furthermore, F-1 is also an isometry.

4.3 Theorem Isometries preserve intrinsic distance: if F: M Æ is anisometry of surfaces in R3, then

for any two points p, q in M.(Here r and are the intrinsic distance functions of M and

respectively.)

Proof. First note that isometries preserve the speed and length of curves.The proof is just like the Euclidean case: If a is a curve segment in M,then = F(a) is a curve segment in with velocity ¢ = F*(a¢). SinceF* preserves dot products, it preserves norms, so || a¢ || = || F*(a¢) || =|| F(a)¢ || = || ¢ ||. Hence

Now, if a runs from p to q in M, its image = F(a) runs from F(p) toF(q) in . Reciprocally, if b is a curve segment in from F(p) to F(q) in

, then F-1(b) runs from p to q in M. We have, in fact, established a one-to-one correspondence between the collection of curve segments usedto define r(p,q) and those used for (F(p),F(q)). But as was shown above,corresponding curves have the same length, so it follows at once that r(p, q) = (F(p), F(q)). ◆r

r

MMM

a

L t dt t dt La

b

a

b

a a a a( ) = ¢( ) = ¢( ) = ( )Ú Ú .

a

aMa

Mr

r rp q p q, ,( ) = ( ) ( )( )F F

M

MM

v v= ( ) =F* ;0

F F* • * •v w v w( ) ( ) =

MM


Thus we may think of an isometry as bending a surface into a differentshape without changing the intrinsic distance between any of its points. Con-sequently, the inhabitants of the surface are not aware of any change at all, fortheir geometric measurements all remain exactly the same.

If there exists an isometry from M to , then these two surfaces are saidto be isometric. For example, if a piece of paper is bent into various shapeswithout creasing or stretching, the resulting surfaces are all isometric (Fig. 6.9).

To study isometries it is convenient to separate the geometric condition ofpreservation of dot products from the one-to-one and onto requirements.

4.4 Definition A local isometry F: M Æ N of surfaces is a mapping thatpreserves dot products of tangent vectors (that is, F* does).

Thus an isometry is a local isometry that is both one-to-one and onto.If F is a local isometry, the earlier argument still shows that F is a regular

mapping. Then for each point p of M the inverse function theorem (5.4 ofChapter 4) asserts that there is a neighborhood U of p in M that F carriesdiffeomorphically onto a neighborhood V of F(p) in N. Now, U and V arethemselves surfaces in R3, and thus the mapping F |U: U Æ V is an isometry.In this sense a local isometry is, indeed, locally an isometry.

There is a simple patch criterion for local isometries using the functions E,F, and G defined in Section 4 of Chapter 5.

4.5 Lemma Let F: M Æ N be a mapping. For each patch x: D Æ M, con-sider the composite mapping

x x= ( ) ÆF D N: .

M


FIG. 6.9

Then F is a local isometry if and only if for each patch x we have

(Here need not be a patch, but , and are defined for it as usual.)

Proof. Suppose the criterion holds—and only for enough patches tocover all of M. Then by one of the equivalences in Exercise 1, to show thatF* preserves dot products we need only prove that

But as we saw in Chapter 4, it follows immediately from the definitionof F* that F*(xu) = u and F*(xv) = v. Thus the equations above followfrom the hypotheses E = , F = , G = . Hence F is a local isometry.

Reversing the argument, we deduce the converse assertion. ◆

This result can sometimes be used to construct local isometrics. In the sim-plest case, suppose that M is the image of a single patch x: D Æ M.

Then if y: D Æ N is a patch in another surface, a mapping F: M Æ N isdefined by

If E = , F = , G = , then by the above criterion, F is a local isometry.

4.6 Example (1) Local isometry of a plane onto a cylinder. The plane R2 may be considered as a surface, with natural frame field U1, U2. If

x: R2 Æ M

is a parametrization of some surface, then Exercise 1 of this section showsthat x is a local isometry if

Since x*(U1) = xu, x*(U2) = xv, and Ui • Uj = dij, this is the same as requir-ing E = 1, F = 0, G = 1.

To take a concrete case, the parametrization

of the cylinder M: x2 + y2 = r2 has E = 1, F = 0, G = 1. Thus x is a localisometry that wraps the plane R2 around the cylinder, with horizontal linesgoing to cross-sectional circles and vertical lines to rulings of the cylinder.

x u v rur

rur

v, , ,( ) = ÊËÁ

ˆ¯cos sin

x x* • * • , .U U U U i ji j i j( ) ( ) = £ £for 1 2

GFE

F u v u v u v Dx y, , for , in( )( ) = ( ) ( ) .

GFExx

x x x x x x x xu u u v u v v vF F F F= ( ) = ( ) ( ) = ( )* , • * • * * .,

GFEx

E E F F G G= = =, , .


(2) Local isometry of a helicoid onto a catenoid. Let H be the helicoid thatis the image of the patch

Furnish the catenoid C with the canonical parametrization y: R2 Æ C dis-cussed in Example 7.4 of Chapter 5. Thus

Let F: H Æ C be the mapping such that

To prove that F is a local isometry, it suffices to check that

F carries the rulings (v constant) of H onto meridians of the surface of rev-olution C, and wraps the helices (u constant) of H around the parallels of C.In particular, the central axis of H (z axis) is wrapped around the minimalcircle x = 0 of C.

Figure 6.10 shows how a sample strip of H is carried over to C.

E E F F G u h G= = = = = + = =1 0 1 2 2, , .

F u v u vx y, ,( )( ) = ( ).

y u v g u h u v h u v

g u u h u u

, , , ,

h ,

( ) = ( ) ( ) ( )( )

( ) = ( ) = +-

cos sin

sin .1 21

x u v u v u v v, , ,( ) = ( )cos sin .


FIG. 6.10

Suppose that the helicoid H (or at least a finite region of it) has beenstamped like an automobile fender out of a flexible sheet of steel—the patchx does this. Then H may be wrapped into the shape of a catenoid with nofurther distortion of the metal (Exercise 5 of Section 5).

A similar experiment may be performed by cutting a hole in a ping-pongball representing a sphere in R3. Mild pressure will then deform the ball intovarious nonround shapes, all of which are isometric. For arbitrary isometricsurfaces M and in R3, however, it is generally not possible to bend M(through a whole family of isometric surfaces) so as to produce .

There are special types of mappings other than (local) isometries that areof interest in geometry.

4.7 Definition A mapping of surfaces F: M Æ N is conformal providedthere exists a real-valued function l > 0 on M such that

for all tangent vectors to M. The function l is called the scale factorof F.

Note that if F is a conformal mapping for which l has constant value 1,F is a local isometry. Thus a conformal mapping is a generalized isometry forwhich lengths of tangent vectors need not be preserved—but at each point pof M the tangent vectors at p all have their lengths stretched by the samefactor.

The criteria in Lemma 4.5 and in Exercise 1 below may easily be adaptedfrom isometries to conformal mappings by introducing the scale factor (orits square). In Lemma 4.5, for example, replace E = by E l2(x) = , andsimilarly for the other two equations.

An essential property of conformal mappings is discussed in Exercise 8.

Exercises

1. If F: M Æ N is a mapping, show that the following conditions on itstangent map at one point p are logically equivalent:

(a) F* preserves inner products.(b) F* preserves lengths of tangent vectors, that is, || F*(v) || = || v || for allv at p.(c) F* preserves frames: If e1, e2 is a tangent frame at p, then

is a tangent frame at F(p).

F F* , *e e1 2( ) ( )

EE

F p p* v p v( ) = ( )l

MM


(d) For some one pair of linearly independent tangent vectors v and w atp,

[Hint: It suffices, for example, to prove (a) fi (c) fi (d ) fi (b) fi (a).]These are general facts from linear algebra; in this context they provide

useful criteria for F to be a local isometry.

2. Show that each of the following conditions is necessary and sufficient forF: M Æ N to be a local isometry.

(a) F preserves the speeds of curves: || F(a)¢ || = || a ¢ || for all curves a in M.(b) F preserves lengths of curves: L(F(a)) = L(a) for all curve segments ain M.

3. Prove that intrinsic distance r for a surface M has the same general prop-erties as Euclidean distance d (Ex. 2 of Sec. 2.1), namely,

(a) Positive definiteness: r(p, q) � 0; r(p, q) = 0 if and only if p = q,(b) Symmetry: r(p, q) = r(q, p),(c) Triangle inequality: r(p, r) � r(p, q) + r(q, r), for all points of M.The only difficult case is r(p, q) = 0 fi p = q, which we postpone to Exer-cise 1 of Section 8.1. (Hint for (c): Piecewise differentiable curves areallowed in the definition of r.)

4. Give an example and a proof to show: Local isometries can shrink butnot increase intrinsic distance.

5. Let a, b: I Æ R3 be unit-speed curves with the same curvature functionk > 0. For simplicity, assume that the parametrization

of the u > 0 tangent surface of a is actually a patch. Find a local isometryfrom this surface to:

(a) The u > 0 tangent surface of b.(b) A region D in the plane.

6. Show that the preceding exercise applies to the u > 0 tangent surface ofa helix, and find the image region D in the plane.

7. Let M be the Euclidean plane R2 with the origin removed. Show that theintrinsic distance from (-1, 0) to (1, 0) is 2, but that every curve joining thesepoints has length strictly greater than 2. (Hint: Ex. 11 of Sec. 2.2.)

8. (a) Modify the conditions in Exercise 1 so that they provide criteria forF to be a conformal mapping.

x u v u vT u,( ) = ( ) + ( )a

F F F F* , * * • * • .v v w w v w v w( ) = ( ) = ( ) ( ) =and


(b) Show that a parametrization x: D Æ M is a conformal mapping if andonly if E = G and F = 0.(c) Prove that a conformal mapping preserves angles in this sense: If J isan angle between v and w at p, then J is also an angle between F*(v) andF*(w) at F(p).

9. If F: M Æ is an isometry, prove that the inverse mapping F -1: Æ M is also an isometry. If F: M Æ N and G: N Æ P are (local)isometries, prove that the composite mapping GF: M Æ P is a (local) isometry.

10. Let x be a parametrization of all of M, a parametrization in N. IfF: M Æ N is a mapping such that F(x(u, v)) = (f(u), g(v)), then

(a) Describe the effect of F on the parameter curves of x.(b) Show that F is a local isometry if and only if

(In the general case, f and g are functions of both u and v, and this crite-rion becomes more complicated.)(c) Find analogous conditions for F to be a conformal mapping.

11. Let M be a surface of revolution, and let F: H Æ M be a local isome-try of the helicoid that (as in Example 4.6) carries rulings to meridians andhelices to parallels. Show that M must be a catenoid. (Hint: Use Ex. 10.)

12. Let M be the image of a patch x with E = 1, F = 0, and G a functionof u only (Gv = 0). If the derivative is bounded, show that there isa local isometry of M into a surface of revolution.

Thus any small enough region in M is isometric to a region in a surface ofrevolution.

13. Let x be the geographical patch in the sphere S of radius r (Example2.2 of Ch. 4). Stretch x in the north-south direction to produce a conformalmapping. Explicitly, let

and determine g such that y is conformal. Find the scale factor of y and thedomain D such that y(D) omits only a semicircle of S. (Mercator’s map ofthe earth derives from y: its inverse is Mercator’s projection.)

14. Show that stereographic projection P: S0 Æ R2 (Example 5.2 of Ch. 4)is conformal, with scale factor

y xu v u g v g, , with ,( ) = ( )( ) ( ) =0 0

d duG /( )

E E f gdfdu

F F f gdfdu

dgdv

G G f gdgdv

= ( )ÊËÁˆ¯ = ( ) = ( )ÊËÁ

ˆ¯, .

2 2

, , , ,

xx

MM


15. Let M be a surface of revolution whose profile curve is not closed,and hence has a one-to-one parametrization. Find a conformal mapping F: M Æ R2 such that meridians go to lines through the origin and parallelsgo to circles centered at the origin.

6.5 Intrinsic Geometry of Surfaces in R3

In Chapter 3 we defined Euclidean geometry to consist of those concepts pre-served by Euclidean isometries. The same definition applies to surfaces: Theintrinsic geometry of M Ã R3 consists of those concepts—called isometricinvariants—that are preserved by all isometries F: M Æ . For example,Theorem 4.3 shows that intrinsic distance is an isometric invariant. We cannow state Gauss’s question (mentioned early in the chapter) more precisely:Which of the properties of a surface M in R3 belong to its intrinsic geometry?The definition of isometry (Definition 4.2) suggests that isometric invariantsmust depend only on the dot product as applied to tangent vectors to M. Butthe shape operator derives from a normal vector field, and the examples inSection 4 show that isometric surfaces in R3 can have quite different shapes.In fact, these examples provide a formal proof that shape operators, princi-pal directions, principal curvatures, and mean curvature definitely do notbelong to the intrinsic geometry of M Ã R3.

To build a systematic theory of intrinsic geometry, we must look back atSection 1 and see how much of our work there is intrinsic to M. Using thedot product only on tangent vectors to M, we can still define a tangent framefield E1, E2 on M. Thus from an adapted frame field we can salvage the twotangent vector fields E1, E2—and hence also their dual 1-forms q1, q2. It issomewhat surprising to find that these completely determine the connectionform w12.

5.1 Lemma The connection form w12 = -w21 is the only 1-form that satisfies the first structural equations

Proof. Apply these equations to the tangent vector fields E1, E2. Sinceqi(Ej) = dij, the definition of wedge product gives

d dq w q q w q1 12 2 2 21 1= Ÿ = Ÿ, .

M

l pp( ) = +

( )1

4

2P.

6.5 Intrinsic Geometry of Surfaces in R3 289

Thus by Lemma 2.1, w12 = -w21 is uniquely determined by q1, q2. ◆

5.2 Remark In fact, this proof shows how to construct w12 = -w21

without the use of Euclidean covariant derivatives (as in Section 1). GivenE1, E2 and thus q1, q2, take the equations in the above proof as the definitionof w12 on E1 and E2. Then the usual linearity condition

makes w12 a 1-form on M, and one can easily check (by reversing the argu-ment above) that w12 = -w21 satisfies the first structural equations.

If F: M Æ is an isometry, we can transfer a tangent frame field E1, E2

on M to a tangent frame field 1, 2 on : For each point q in there isa unique point p in M such that F(p) = q. Then define

(Fig. 6.11).

In practice, we often abbreviate these formulas somewhat casually to

Because F* preserves dot products, 1, 2 is a frame field on , since

E E F E F E E Ei j i j i j ij• * • * • .= ( ) ( ) = = d

MEE

E F E E F E1 1 2 2= ( ) = ( )* , * .

E F E

E F E

1 1

2 2

q p

q p

( ) = ( )( )( ) = ( )( )

* ,

*

MMEEM

w w w w12 12 1 1 2 2 1 12 1 2 12 2V v E v E v E v E( ) = +( ) = ( ) + ( )

w q

w w q

12 1 1 1 2

12 2 21 2 2 1 2

E d E E

E E d E E

( ) = ( )( ) = - ( ) = ( )

, ,

, .


FIG. 6.11

5.3 Lemma Let F: M Æ be an isometry, and let E1, E2 be a tangent frame field on M. If 1, 2 is the transferred frame field on ,then

(1) q1 = F*( 1), q2 = F*( 2);(2) w12 = F*( 12).

Proof. (1) It suffices to prove that qi and F*( i) have the same value onE1 and E2. But for 1 � i, j � 2 we have

(2) Consider the structural equation d 1 = 12 Ÿ 2 on . If we applyF*, then by the results in Chapter 4, Section 5,

Hence, by (1), we have

The other structural equation,

gives a corresponding equation, so

But now (2) is an immediate consequence of the uniqueness property(Lemma 5.1), since

◆

From this rather routine lemma we easily derive a proof of the celebratedtheorema egregium of Gauss.

5.4 Theorem Gaussian curvature is an isometric invariant. Explicitly, ifF: M Æ is an isometry, then

for every point p in M.

K K Fp p( ) = ( )( )

M

F F F* * * .w w w21 12 12( ) = -( ) = - ( )

d F

d F

q w q

q w q

1 12 2

2 21 1

= ( ) Ÿ

= ( ) Ÿ

*

* .

,

dq w q2 21 1= Ÿ ,

d Fq w q1 12 2= ( ) Ÿ* .

d F F d F F* * * * .q q w q1 1 12 2( ) = ( ) = ( ) Ÿ ( )

Mqwq

F E F E E Ei j i j i j ij i j* * .q q q d q( )( ) = ( ) = ( ) = = ( )

q

wqq

MEEM


Proof. For an arbitrary point p of M, pick a tangent frame field E1, E2

on some neighborhood of p and transfer via F* to 1, 2 on . By theprevious lemma, F*( 12) = w12. According to Corollary 2.3, we have

Apply F* to this equation. By the results in Chapter 4, Section 5, we get

where F*( ) is simply the composite function (F ). Thus by the preced-ing lemma,

Comparison with dw12 = -Kq1 Ÿ q2 yields K = (F ); hence, in particular,K(p) = (F(p)). ◆

Gauss’s theorem is one of the great discoveries of nineteenth-century mathematics, and we shall see in the next chapter that its implications are far-reaching. The essential step in the proof is the second structural equation

Once we prove Lemma 5.1, all the ingredients of this equation, except K, areknown to derive from M alone—thus K must also. This means that the inhabitants of M Ã R3 can determine the Gaussian curvature of their surfaceeven though they cannot generally find S and have no conception of the shapeof M in R3.

This remarkable situation is perhaps best illustrated by the formula K = k1k2: An isometry need not preserve the principal curvatures, nor theirsum, but it must preserve their product. Thus the shapes that isometric surfacesmay have—although possibly quite different—are by no means unrelated.

A local isometry is, as we have shown, an isometry on all sufficiently smallneighborhoods. Thus it follows from Theorem 5.4 that local isometries pre-serve Gaussian curvature. For example, in Example 4.6 the plane and the cylin-der both have K = 0. (This is why we did not hesitate to call the curvedcylinder “flat.” Intrinsically it is as flat as a plane.) In the second part ofExample 4.6, at corresponding points

both the helicoid and catenoid have Gaussian curvature: (seeExamples 4.3 and 7.4 of Chapter 5).

Gauss’s theorema egregium can obviously be used to show that given sur-faces are not isometric. For example, there can be no isometry of the sphere

- +( )1 1 2 2/ u

x x yu v F u v u v, and , , ,( ) ( )( ) = ( )

d Kw q q12 1 2= - Ÿ .

KK

d K Fw q q12 1 2= - ( ) Ÿ .

KK

d F F d F K F F* * * * *w w q q12 12 1 2( ) = ( ) = - ( ) ( ) Ÿ ( ),

d Kw q q12 1 2= - Ÿ .

wMEE


S (or even a very small region of it) onto part of the plane, since their Gaussian curvatures are different. This is the dilemma of the mapmaker:The intrinsic geometry of the earth’s surface is misrepresented by any flatmap.

The next section is computational; thereafter we will find more isometricinvariants.

Exercises

1. Geodesics belong to intrinsic geometry: In fact, if a is a geodesic in Mand F: M Æ N is a (local) isometry, then F(a) is a geodesic of N. (Hint:Ex. 1 of Sec. 1)

2. Use Exercise 1 to derive the geodesics of the circular cylinder. General-ize to an arbitrary cylinder.

3. For a connected surface, the values of its Gaussian curvature fill an inter-val. If there exists a local isometry of M onto N (in particular, if M and Nare isometric), show that M and N have the same curvature interval. Give anexample to show that the converse is false.

4. Prove that no two of the following surfaces are isometric: sphere, torus,helicoid, cylinder, saddle surface.

5. Bending of the helicoid into the catenoid (4.6). For each number t in theinterval , let xt: R2 Æ R3 be the mapping such that

where C = coshu, S = sinhu, c = cosv, and s = sinv.Now x0 is a patch covering the helicoid, and is a parametrization of

the catenoid—these are mild variants of our usual parametrizations, and thecatenoid now has the z axis as its axis of rotation. If we imagine t to be thetime, then xt for describes a bending of the helicoid M0 thatcarries it onto the catenoid through a whole family of intermediate sur-faces Mt = xt(R2). Prove:

(a) Mt is a surface. (Show merely that xt is regular.)(b) Mt is isometric to the helicoid M0 if . (Show that Ft: M0 Æ Mt

is an isometry, where

F u v u vt tx x0 , .( )( ) = ( ),

t < p /2

Mp 2

0 2� �t p /

xp 2

x t u v t Sc Ss v t Cs Cc u, , , , , ,( ) = ( ) + -( )cos sin

0 2� �t p /


Also show that for , is local isometry.)(c) Each Mt is a minimal surface. (Compute xuu + xvv = 0.)(d) Unit normals are parallel on orbits. Along the curve t Æ xt(u, v) by whichthe point x0(u, v) of M0 moves to , the unit normals Ut of successivesurfaces are parallel.(e) Gaussian curvature is constant on orbits. Find Kt(xt(u, v)), where Kt isthe Gaussian curvature of Mt.A brilliant series of illustrations of this bending is given in Struik [S]. Theyare not diminished by the following computer tour de force.

6. (Computer continuation.) (a) Plot the surface Mt for at least six valuesof t from t = 0 (helicoid) to (catenoid).(b) Animate the series of plots in (a).

7. Show that every local isometry of the helicoid H to the catenoid C mustcarry the axis of H to the central circle of C, and the rulings of H to themeridians of C, as in Example 4.6.

6.6 Orthogonal Coordinates

We have seen that the intrinsic geometry of a surface M Ã R3 may beexpressed in terms of the dual forms q1, q2 and connection form w12 derivedfrom a tangent frame field E1, E2. These forms satisfy—

the first structural equations:

the second structural equation:

In this section we develop a practical way to compute these forms—and hencea new way to find the Gaussian curvature of M.

The starting point is an orthogonal coordinate patch x: D Æ M, one forwhich F = xu • xv = 0. Since xu and xv are orthogonal, dividing by theirlengths will produce frames.

6.1 Definition The associated frame field E1, E2 of an orthogonal patchx: D Æ M consists of the orthogonal unit vector fields E1 and E2 whose valuesat each point x(u, v) of x(D) are

x xu vE G= =and

d Kw q q12 1 2= - Ÿ .

d

d

q w q

q w q

1 12 2

2 21 1

= Ÿ

= Ÿ

,

;

t < p /2

Mp 2

Fp 2t < p /2


In Exercise 7 of Section 4.4, we associated with each patch x the coordi-nate functions u and v, which assign to each point x(u, v) the numbers u andv, respectively. For example, for the geographical patch x of Example 2.2 ofChapter 4, the coordinate functions are the longitude and latitude functionson the sphere S. In the extreme case when x is the identity map of R2,the coordinate functions are just the natural coordinate functions (u, v) Æ u,(u, v) Æ v on R2.

For an orthogonal patch x with associated frame field E1, E2, we shallexpress q1, q2, and w12 in terms of the coordinate functions u, v. Since x isfixed throughout the discussion, we shall run the risk of omitting the inversemapping x-1 from the notation. With this convention, the coordinate func-tions u = u(x-1) and v = v(x-1) are written simply u and v, and similarly xu

and xv now become tangent vector fields on M itself. Thus the associatedframe field of x has the concise expression

(1)

The dual forms q1, q2 are characterized by qi(Ej) = dij, and in the exercisereferred to above it is shown that

Thus we deduce from (1) that

(2)

By using the structural equations, we find analogous formulas for w12 andK. Recall that for a function f, df = fu du + fv dv, where the subscripts indicatepartial derivatives. Hence

where we have used the alternation rule for wedge products and substitutedand from (2). Comparison with the first structural

equations dq1 = w12 Ÿ q2 and dq2 = -w12 Ÿ q1 shows thatdu E= q1 /dv G= q 2 /

d d E du E dv du EG

du

d d G dv G du dv GE

dv

vv

uu

q q

q q

1 2

2 1

= ( ) Ÿ = ( ) =-( )

Ÿ

= ( ) Ÿ = ( ) =-( )

Ÿ

,

,

q q1 2= =E du Gdv, .

du dv

du dv

u u

v v

x x

x x

( ) = ( ) =

( ) = ( ) =

1 0

0 1

, ,

, .

EE

EG

u v1 2= =

x x, .

x xu vu v

E u v

u v

G u v

,

,and

,

,

( )( )

( )( )

.

6.6 Orthogonal Coordinates 295

(3)

The logic is simple: By the computations above, this form satisfies the firststructural equations; hence by uniqueness (Lemma 5.1), it must be w12.

6.2 Example Geographical coordinates on the sphere. For the geographi-cal patch x in the sphere S (Example 2.2 of Chapter 4), we found E =r2 cos2 v, F = 0, G = r2. Thus by formula (2) above,

Now and ; hence by (3),

The associated frame field of this patch is the same one obtained inExample 1.6 from the spherical frame field in R3. With the notational shift u Æ J, v Æ j, the forms above are (necessarily) also the same. But now wehave a simple way to compute them directly in terms of the surface with noappeal to the geometry of R3.

Finally, we derive a new expression for the Gaussian curvature. In thiscontext, exterior differentiation of w12 as given in (3) yields

From (2) we get

hence

Thus the formula above becomes

Now compare this with the second structural equation, dw12 = -Kq1 Ÿ q2. ◆

6.3 Proposition If x: D Æ M is an orthogonal patch, the Gaussian cur-vature K is given in terms of x by

dEG

GE G

Eu

u

v

v

w q q12 1 2

1=

( )ÊËÁ

ˆ¯ +

( )ÊËÁ

ˆ¯

ÏÌÓ

¸˝˛

Ÿ

.

- = = Ÿdv du du dvEG

11 2q q .

q q1 2Ÿ = EG du dv;

d EG

dvduE

dudvGv

v

u

u

w12 = -( )Ê

ËÁˆ¯ +

( )ÊËÁ

ˆ¯

w12 = sin .v du

G u ( ) = 0E r vv sin( ) = -

q q1 2= =r v du r dvcos .,

w12 =-( )

+( )E

Gdu

EdvGv u

.


This is perhaps the most elegant of the dozens of formulas that have beenfound for Gaussian curvature. By contrast with the formula in Corollary 4.1of Chapter 5, the functions , , (which describe the shape operator) nolonger appear. Indeed, since K is now expressed solely in terms of E, F, G,by using Lemma 4.5 we get another proof of the isometric invariance ofGaussian curvature.

Exercises

1. Compute the dual 1-forms, connection form w12, and Gaussian curva-ture for the associated frame field of the following orthogonal patches:

(a) x(u, v) = (u cosv, u sinv, bv), helicoid.(b) x(u, v) = (u cosv, u sinv, u2/2), paraboloid of revolution.(c) x(u, v) = (u cosv, u sinv, au), cone.

2. A parametrization x: D Æ M is isothermal provided E = G and F = 0.(Thus a fine network of parameter curves cuts M into small regions that arealmost square.) By Exercise 8 of Section 4, x is a conformal mapping withscale function l such that E = G = l2. Prove:

(a) where D is the Laplacian: Df = fuu + fvv.

(b) The mean curvature H is zero if and only if xuu + xvv = 0.

3. If x is a principal patch (Ex. 8 of Sec. 5.4), prove

(a)

(b)

(Hints: For (a), compare Cor. 1.5 with Ex. 8 of Sec. 5.4. For (b), use (a) andthe Codazzi equations from Thm. 1.7.)

6.7 Integration and Orientation

A primary goal of this section is to define the integral of a 2-form over acompact oriented surface. This notion does not involve geometry at all; it

L Nv v u uHE HG= =, .

w w13 23= =L N

Edu

Gdv, .

K E E= - = - ( )D Dlog / log / ,l l2 2

KEG

GE G

Eu

u

v

v

=- ( )Ê

ËÁˆ¯ +

( )ÊËÁ

ˆ¯

ÏÌÓ

¸˝˛

1 .

6.7 Integration and Orientation 297

belongs to the integral calculus on surfaces (Chapter 4, Section 6). However,we motivate the definition by considering some geometric applications. Ori-entation will be involved, for its connection with integration is already shown

by the elementary calculus formula

Perhaps the simplest use of double integration in geometry is in finding thearea of a surface. To discover a proper definition of area, we start with apatch x: D Æ M and ask what the area of its image should be. Let DR be asmall coordinate rectangle in D with sides Du and Dv. Now, x distorts DR intoa small curved region x(DR) in M, marked off by four segments of parame-ter curves, as shown in Fig. 6.12.

We have seen that the segment from x(u, v) to x(u + Du, v) is linearly approximated by the vector Du xu evaluated at (u, v), and the segment from x(u, v) to x(u, v +Dv) is approximated by Dv xv. Thus the region x(DR) is approximated by the parallelogram in the tangent plane at x(u, v) that has these vectors as its sides. From Chapter 2, Section 1 we know that thisparallelogram has area

We conclude that the area of x(DR) should be approximately times the area DuDv of DR. So at each point of D, the familiar expression

gives the rate at which x is expanding area. Thus it would benatural to define the area of the whole region x(D) to be

But since D is an open set, such integrals may well be improper. To avoid thiswe must modify the definition of coordinate patch.

7.1 Definition The interior R° of a rectangle R: a � u � b, c � v � d isthe open set a < u < b, c < v < d. A 2-segment x: R Æ M is patchlike providedthe restricted mapping x: R° Æ M is a patch in M.

DED F du dv-ÚÚ 2 .

EG F- 2

EG F- 2

D D D D D Du v u v EG F u vu v u vx x x x¥ = ¥ = - 2 .

f x dx f x dxb

a

a

b( ) ( )Ú Ú= - .


FIG. 6.12

The continuous function is bounded on the closed rectangleR, so the area of a patchlike 2-segment is well defined and finite, since in thedouble integral above the open set, D will be replaced by R.

A patchlike 2-segment x: R Æ M is not required to be one-to-one on theboundary of R, so its image may not be very rectangular.

7.2 Example Areas of surfaces. We begin with two familiar cases.(1) The sphere of radius r. If the formula defining the geographical patch

is applied to the closed rectangle R: the resultis a 2-segment covering the whole sphere. Since

we get = r2 cos v. Thus the area of the sphere is

a result well known to Euclid.(2) Torus of radii R > r > 0. From Example 2.5 of Chapter 4 we derive a

patchlike 2-segment covering the torus. Here

with -p � u, v � p, so the area of the torus is

(3) The bugle surface (Example 7.6 of Chapter 5). Every surface of revo-lution M has a canonical parametrization with E = 1, F = 0, G = h2. On therectangle R: a � u � b, 0 � v � p, x is a patchlike 2-segment whose imageis the closed region Zab of M between the parallels u = a and u = b (Fig. 6.13).Thus the area of the zone Zab is

In the case of the bugle surface, h(u) = ce-u/c. Hence

To find the area of the entire bugle—a noncompact surface—let Zab expand,with a Æ 0 and b Æ •. The corresponding areas are positive and increasing,and hence approach a limit. The limit of Aab as a Æ 0 and b Æ • is 2pc2.Hence, in particular, the bugle has finite area. ◆

A c e du c e eabu c

a

ba c b c= = -( )- - -Ú2 2 2p p .

A h duaba

b

a

b

h du dv= = ÚÚÚ 20

2

pp

.

A r R r u du dv Rr= +( ) =-- ÚÚ cos .

p

p

p

p

p4 2

EG F r R r u- = +( )2 cos ,

A r v du dv r= =-- ÚÚ 2 24

2

2

cos ,pp

p

p

p

EG F- 2

E r v F G r= =2 2 0cos , , = ,2

- -p p p p� � � �u v, / / ,2 2

EG F- 2


To define the area of a complicated region, we follow the usual schemefrom elementary calculus: Break the region into simple pieces and add their areas.

7.3 Definition A paving of a region P in a surface M consists of a finitenumber of patchlike 2-segments x1, . . . , xk whose images fill M in such a waythat each point of M is in at most one set x i(Ri

o).

In short, the images of the segments cover all of P, overlapping only ontheir boundaries (see Fig. 6.14).

Not every region is pavable. Since pavings are finite, compactness is cer-tainly necessary (Lemma 7.2 in Chapter 4). In fact, a compact region ispavable if its boundary consists of a finite number of regular curve segments.In particular, an entire compact surface is always pavable. The area of a


FIG. 6.13

FIG. 6.14

pavable region is defined to be the sum of the areas of x1(R1), . . . , xk(Rk) forany paving of P .

The preceding exposition shows that a treatment of area does not requiredifferential forms, but integration of 2-forms will give area and much morebesides (see Remark 7.8). The first question is this: Which 2-form should be integrated over a patchlike 2-segment to get the area of its image? By Definition 6.3 of Chapter 4,

Thus we need a 2-form whose value on xu, xv is

This suggests

7.4 Definition An area form on a surface M is a differentiable 2-form mwhose value on any pair of tangent vectors is

(The sign ambiguity cannot be avoided.)

Because forms are bilinear, it would be equivalent to require that m(E1, E2) = ±1 for every frame E1, E2 on M.

7.5 Lemma A surface M has an area form if and only if it is orientable.On a connected orientable surface there are exactly two area forms, whichare negatives of each other. (These are denoted by ±dM.)

In fact, by Definition 7.4 of Chapter 4, M is orientable if and only if thereis a nonvanishing 2-form on it. So if M has an area form, it is certainly ori-entable. The remainder of the proof follows the same pattern as for Propo-sition 7.5 in Chapter 4. Indeed, for M Ã R3 there is a natural one-to-onecorrespondence between unit normals U and area forms dM given by

To orient a connected orientable surface M is to choose one of its two areaforms—or equivalently, one of its two unit normals.

Finding area is not a typical integration problem, because area is alwayspositive. Thus to find area by integrating an area form dM we have to be

dM Uv w v w,( ) = ± ¥• .

m v w v v w w v w v w,( ) = ± ( )( ) - ( )( ) = ± ¥• • • .2 1 2

x xu v EG F¥ = - 2 .

m m= ( )ÚÚ ÚÚxu v

Rdu dvx x, .


careful about signs. Let x be a patchlike 2-segment in a surface oriented byarea form dM. By definition,

Now there are two cases:(1) If dM(xu, xv) > 0, we say that x is positively oriented. Then by the def-

inition of area form,

hence is the area of x(R).

(2) If dM(xu, xv) < 0, we say that x is negatively oriented. Then

hence is minus the area of x(R).Thus, to find the area of a pavable region P by integrating the area form,

we cannot use an arbitrary paving. The paving must be positively oriented,that is, consist only of positively oriented patchlike 2-segments. Then

Now we replace the area form by an arbitrary 2-form to get the definitionwe are looking for.

7.6 Definition Let v be a 2-form on a pavable oriented region P in asurface. The integral of v over P is

where x1, . . . , xk is a positively oriented paving of P.

There is a consistency problem with this definition that appears already inelementary calculus. Any two choices of paving for P must produce the samevalue for the integral. A formal proof of this is not elementary and belongsproperly to analysis rather than to geometry.

Area forms make it easy to describe integration of functions. If f is a continuous function on a pavable region P, then its integral over P is defined

to be . Evidently, this is an analogue of the usual integral of elementary calculus, since dx Ÿ dy is the area form of the Euclidean plane.

f dx dyÚÚf dMP

ÚÚ

v vii

=ÚÚ ÚÚÂP x

,

area areaP( ) = ( )( ) =Â ÂÚÚxx

i ii ii

R dM.

dMxÚÚ

dM EG Fu vx x,( ) = - - 2 ;

dMxÚÚ

dM EG Fu vx x,( ) = - 2 ;

dM dM du dvu vR

= ( )ÚÚ ÚÚxx x, .


7.7 Remark Improper integrals. We have defined integration only overcompact regions; however, a positive function f > 0 can be integrated over anarbitrary surface M (or open region) by defining its integral to be the leastupper bound of the integrals of f over all pavable regions P in M:

This becomes +• if no upper bound exists. Setting f = 1 gives the area of M,as in Example 7.2(3). (For f < 0, switch to greatest lower bounds and -•.)

7.8 Remark Geometry uses two related notions of integration, distin-guished by the effect of change of variables. Arc length, area, and even theintegral of a function belong to absolute integration (or measure theory),which is independent of parametrization and involves the absolute value ofthe relevant Jacobian. By contrast, integration of differential forms dependson orientation and involves the signed Jacobian. This version has a crucialadvantage: Stokes theorem (Theorem 6.5 of Chapter 4) and its many consequences.

Exercises

1. For a Monge patch x(u, v) = (u, v, f(u, v)), show that the area of x(D) is given by the usual formula from elementary calculus. Deduce that A(x(D)) � A(D).

2. (Theorem of Pappus.) Let M be a surface of revolution whose profilecurve has finite length L. Find a formula for the area of M and interpret it as A = 2p L, where is the average distance of M from the axis ofrevolution.

3. Let x be the usual parametrization of the torus of revolution T (Example2.5 of Ch. 4), with T oriented by the outward unit normal U. Compute the

integral where in each case, v is the 2-form on T such that:

(a) v(xu, xv) is the square of the distance from x(u, v) to the origin 0 in R3.(b) v(xu, xv) = U • xu ¥ xv.

4. Let M be a compact surface oriented by dM, and let -M be the samesurface oriented by -dM. Prove

(a) (c1,c2 constant).

(b) (Hint: Use Ex. 5.)v vM M-ÚÚ ÚÚ= - .

c v c v c v c vM M M

1 2 2 1 1 2 21 +( ) = +ÚÚ ÚÚ ÚÚ

vTÚÚ

hh

f dM dMMÚÚ ÚÚ= lub .

P


(c)

(d) If f � g, then

(Note the effect of f = 0 or g = 0.)

5. If F: M Æ N is an orientation-preserving diffeomorphism of compactoriented surfaces, show that

for any 2-form v on N. (Hint: Ex. 7 of Sec. 4.6.)

6. A diffeomorphism F: M Æ N is area-preserving provided the area of anypavable region P in M is the same as the area of its image F(P ) in N. Prove:

(a) A diffeomorphism F: M Æ N is area-preserving if and only if

for all patchlike 2-segments x in M with = F(x) in N. (Here “all” can bereplaced by “sufficiently many to cover M.”)(b) Isometries are area-preserving, isometric surfaces have the same area(include the noncompact case).(c) The cylindrical projection in Example 5.2(1) of Chapter 4 is area-preserving but not an isometry. (Deduce the standard formula for the area of a zone in the sphere.)(d) A mapping F: M Æ N is an isometry if and only if it is conformal andarea-preserving.

6.8 Total Curvature

One of the most important geometric invariants of a surface is gotten by inte-gration of curvature.

8.1 Definition Let K be the Gaussian curvature of a compact surface Moriented by area form dM. Then

is the total Gaussian curvature of M.

The same definition applies to any pavable region in M.To compute total curvature of M we add the total curvatures of each

patchlike 2-segment x of a paving of M. With the usual notation for thedomain R of x,

K dMMÚÚ

x

EG F EG F- = -2 2

F v vM N

* ( ) =ÚÚ ÚÚ

f dM g dMM MÚÚ ÚÚ£ .

f dM f dMM MÚÚ ÚÚ= -( )

-.


As before, K(x) can be computed explicitly using Corollary 4.1 of Chapter 5or by Proposition 6.3.

8.2 Example Total curvature of some surfaces.(1) Constant curvature. If the Gaussian curvature of M is constant, then

the total curvature of M is

Thus a sphere of radius r has total curvature , and the bugle

surface has total curvature .(2) Torus. Let x be the 2-segment in Example 7.2 that covers the torus T.

Then the area form dT has coordinate expression

In Example 7.1 of Chapter 5 we computed

Hence the torus has total Gaussian curvature

Thus the negative curvature of the inner half of the torus has exactly bal-anced the positive curvature of its outer half.

(3) Catenoid. This surface is not compact, and its area is infinite; never-theless, its total curvature is finite. When the parametrization in Example 7.1of Chapter 5 is restricted to the rectangle R: -a � u � a, 0 � v � 2p, itbecomes a patch-like 2-segment covering the zone Za between the parallels u = -a and u = +a (Fig. 6.15). From this example we get

and . Hence the zone has total curvatureEG c u c= ( )cosh /2

Kc u c

x( ) =-

( )1

2 4cosh /

K dT u du dvT

= =ÚÚ ÚÚ --cos .0

p

p

p

p

Ku

r R r ux( ) =

+( )cos

cos.

x* cos .dT EG F du dv r R r u du dv( ) = - = +( )2

- = -( )( )2 1 22 2p p/c c

4 1 42 2p p= ( )( )/r r

K dM K dM KA MM M

= = ( )ÚÚ ÚÚ .

K dM K dM K dM

K EG F du dv

R R

c

d

a

b

xx x x

x

ÚÚ ÚÚ ÚÚÚÚ

= ( ) = ( ) ( )

= ( ) -

* *

.2

6.8 Total Curvature 305

As a Æ •, Za expands to fill the whole surface. So the total curvature of thecatenoid is

The total curvatures computed above are all multiples of 2p, and nonedepends on the particular parameters (radius r, constant c, . . .) of its surface.This is no accident; a major reason appears in Section 6 of the next chapter,but see also Exercise 6 of this section. ◆

8.3 Definition Let M and N be surfaces oriented by area forms dM anddN. Then the Jacobian of a mapping F: M Æ N is the real-valued functionJF on M such that

The Jacobian has considerable geometric significance. Let v, w be tangentvectors to M at a point p. Then

(*)

Suppose F is regular at p. Then if vectors v and w at p are independent, soare their images F*v and F*w at F(p). Thus dM(v, w) and dN(F*(v), F*(w))are both nonzero; hence JF(p) π 0. Evidently the converse is also true, so Fis regular at p if and only if JF (p) π 0.

The sign of JF (p) π 0 is also informative, because if dM(v, w) > 0, thenJF (p) and dN(F*(v), F*(w)) have the same sign. Thus F is

J dM F dN dN F FF p v w v w v w( ) ( ) = ( )( ) = ( ) ( )( ), , ,* * * .

F dN J dMF* .( ) =

K dMacM a

= - ÊËÁ

ˆ¯ = -ÚÚ Æ•

4 4p plim tanh .

K dMu c

du dvaca

a

= - ( ) = - ÊËÁ

ˆ¯ÚÚ ÚÚ-x

14

20

2

cosh /tanh .

p

p


FIG. 6.15

orientation-preserving at p if JF (p) > 0, andorientation-reversing at p if JF (p) < 0.

Furthermore, taking absolute values in (*) gives

(**)

The discussion of area in Section 7 shows that the area of a small regionDM in M marked off by “short” vectors v and w is approximately |dM(v, w)|.Since |dN(F*(v), F*(w)| approximates the area of the image region F(DM)(Fig. 6.16), we interpret (**) as

Thus |JF(p)| is the rate at which F is expanding area at p.In view of the pointwise definitions above, a mapping F: M Æ N of

oriented surfaces is said to be orientation-preserving if JF > 0, orientation-reversing if JF < 0.

For preservation of area, see Exercise 6 of Section 7. A signed version ofarea contains more information. We call

the algebraic area of F(M). The discussion above shows that roughly speak-ing, each small region DM in M contributes to this total the signed area ofits image F(DM) in N, the sign being

(1) Positive, if the orientation of F(DM) agrees with that of N;(2) Negative, if these orientations disagree (so F has turned DM over);(3) Zero, if F collapses DM to a curve or point.

Let us consider what this means in the case of the Gauss map G of an ori-ented surface M in R3. As defined in Exercise 4 of Section 5, G maps M tothe unit sphere S by moving the selected unit normal U(p) at p in M, bydistant parallelism, to the origin of R3—there it points to G(p) in S (see Fig. 6.17).

J dM F dNFM M

= ( )ÚÚ ÚÚ *

J M F MF p( ) ( ) ª ( )area of area ofD D .

J dM dN F FF p v w v w( ) ( ) = ( ), ,* * .


FIG. 6.16

8.4 Theorem The Gaussian curvature K of an oriented surface M Ã R3

is the Jacobian of its Gauss map.

(Here the unit sphere is oriented, as usual, by its outward normal or thecorresponding area form dS.)

Proof. If U = SgiUi is the unit normal orienting M, then the Gauss mapis G = (g1, g2, g3). Note that if S is the shape operator determined by U,then

and by Proposition 7.5 of Chapter 1,

Hence G*(v) and -S(v) are parallel for any tangent vector v to M, as sug-gested in Fig. 6.17. (Recall that parallel here means having the same naturalEuclidean coordinates.)

To prove the theorem we must show that

so we evaluate these 2-forms on an arbitrary pair of tangent vectors to M.Lemma 3.4 of Chapter 5 gives

On the other hand,

A triple scalar product depends only on the Euclidean coordinates of itsvectors, so we can replace G*(v), G*(w) by the vectors -S(v), -S(w) parallel

G d d G G U G G G* , * , * • * * .S S( )( ) = ( ) ( )( ) = ( )( ) ( ) ¥ ( )v w v w p v w

K dM K dM K U

U S S

( )( ) = ( ) ( ) = ( ) ( ) ¥

= ( ) ( ) ¥ ( )v w p v w p p v w

p v w

, , •

• .

K dM G d= ( )* S ,

G g U Gi i* .v v p( ) = [ ] ( )( )Â

- ( ) = — = [ ] ( )ÂS U g Uv i iv v p ,

U


FIG. 6.17

to them. Furthermore, by the definition of G and the special character ofthe unit sphere S, the vectors U(p) and (G(p)) are also parallel (Fig. 6.17).

Thus the two triple scalar products above are the same—and the proofis complete. ◆

8.5 Corollary The total Gaussian curvature of an oriented surface M Ã R3 equals the algebraic area of the image of its Gauss map G: M Æ S.

To prove this it suffices to integrate the form KdM = G*(dS) over M.The following readily proved special case is easier to use since it involves

only ordinary area.

8.6 Corollary Let R be an oriented region in M Ã R3 on which(1) the Gauss map G is one-to-one (the values of U at different points are

never parallel), and(2) either K ≥ 0 or K £ 0 (K does not change sign).

Then the total curvature of R is ±area of G(R), where the sign is that of K.(Evidently, this area does not exceed 4p.)

For example, consider the torus T as in Fig. 5.21. Its Gauss map G sendsthe outer half O of T (where K � 0) in one-to-one fashion onto the wholesphere S—and does the same for the inner half I (where K � 0). Thus thetorus has total curvature A(S) - A(S) = 0, as found in Example 8.2(2) by anexplicit integration.

Next, consider the catenoid C, with rotation axis the x axis as in Fig. 6.15.Considering the normal vectors to the profile curve showsthat the Gauss map G carries this curve in one-to-one fashion onto an opensemicircle of S. Then, considering U on the parallels of C shows that G is aone-to-one map onto S—omitting only the two points ±(1, 0, 0). Thus thetotal curvature of the catenoid is -4p, as computed analytically in Example8.2(3).

An unexpected application of total Gaussian curvature is to curves, whereit provides a proof of the main assertion of Fenchel’s Theorem (Exercise 18in Section 2.4), namely,

Proof. Let M be a tube around a, with parametrization x as in Exercise 17 of Section 5.4. This exercise shows that the region B in M onwhich Gauss curvature is nonnegative is x(D), where D is the rectangle 0 � u � L, and L is the length of a.p p/ /2 3 2� �v

A simple closed curve in has total curvature dsa k pa

R3 2Ú ≥ .

y c x c= ( )cosh /

U


Using the formulas for K and W in that exercise, we find the remark-able result

(*)

Next, we show geometrically that the total curvature of B is at least 4p.Let II be the Euclidean tangent plane to the unit sphere S Ã R3 at a pointq, so II is the plane through q orthogonal to the vector from 0 to q. Takea distant plane parallel to II and move it parallel to itself until it reachestangency to M—at, say, the point p (Fig. 6.18). This plane II¢ is tangent toT, and M lies entirely on one side of it. Thus, of the curvature choices inRemark 3.3 of Chapter 5, only K(p) < 0 is ruled out—hence K(p) � 0.

Evidently, the Gauss map G of M sends p to q in S. Thus G maps Bonto S. Then K(p) � 0 implies

In combination with (*), this proves the result. ◆

On an oriented surface the ambiguity in the measurement of angles men-tioned in Section 1 of Chapter 2 can be decisively reduced. Let M be a surfaceoriented by the unit normal U. If v is a vector tangent to M, then propertiesof the cross product show that the tangent vector

is orthogonal to v and has the same length. So J is a linear operator on eachtangent space to M that rotates each vector through +90°. We call J the rotation operator of M.

8.7 Definition Let v and w be unit tangent vectors at a point of an ori-ented surface M. A number j is an oriented angle from v to w provided

J Uv v( ) = ¥

K dMB

≥ÚÚ 4p.

K dT ds v dv dsB

L L

= - =ÚÚ Ú Ú Úk kp

p

0 2

3 2

02cos .


FIG. 6.18

There is a unique choice of j in the interval 0 � j < 2p. Then every ori-ented angle from v to w is given by j + 2pk for some integer k. The samescheme gives angles for any pair of nonzero tangent vectors: Simply divideeach by its length to produce unit vectors.

8.8 Lemma Let a: I Æ M be a curve in an oriented surface M. If Vand W are nonvanishing tangent vector fields on a, there is a differentiablefunction j on I such that for each t in I, j(t) is an oriented angle from V(t)to W(t).

Proof. As just mentioned, angles are unchanged if V and W are reducedto unit vector fields. Then V, J(V) is a frame field on a. Orthonormalexpansion gives W = fV + gJ(V), where

Since 1 = W • W = f 2 + g2, we can apply Exercise 12 of Section 2.1 to geta differentiable function j on I such that

Evidently, j is the required function. ◆

We call j an angle function from V to W, and sometimes write j = –(V, W). The exercise mentioned in the proof also implies that j isuniquely determined by its value at any one t0 Œ I. Thus any two choices ofj differ by a constant integer multiple of 2p. Many applications of anglefunctions will appear later on.

Consistency is the essence of orientability. In studying a surface Moriented by an area form dM, we prefer using positively oriented patches,dM(xu, xv) > 0 and positively oriented frame fields, dM(E1, E2) = +1. (Recallthat the only possible values of an area form on a frame are ±1.)

When a frame field E1, E2 on M is positively oriented, then the wedgeproduct of its dual 1-forms is precisely the selected area form:

To prove this, it suffices to note that both sides have the value +1 on E1, E2.A minus sign appears in this equation if the frame field is negatively oriented.

The properties of J show that if u is a unit tangent vector, then u, J(u) isa frame—in fact, a positively oriented frame. Thus any nonvanishing vectorfield V on M determines a positively oriented frame field, namely,

dM = Ÿq q1 2 .

f g= cos j j, = sin .

f W V g W J V= = ( )• • .and

w v v= + ( )cos sin .j j J


We call this the associated frame field of V.

Exercises

1. Using the natural orientation of R2 given by the area form du Ÿ dv, prove:(a) For a mapping F = (f, g): R2 Æ R2, the definition of Jacobian in thetext gives the usual formula J = fugv - fvgu.(b) The Jacobian of a patch x: D Æ M in an oriented surface M is

, where the sign depends on whether x is positively or nega-tively oriented.

2. Let D be a small disk in M centered at p, and let a parametrize its bound-ary ∂D. By drawing a few unit normals to M along a, sketch the image of aunder the Gauss map of M if:

(a) p is any point of an ellipsoid M.(b) p is any point of a circular cylinder.(c) p is the origin in M: z = xy.

3. (a) Prove that the Gauss map of a surface M Ã R3 is conformal if andonly if its principal curvature functions satisfy k2

1 = k22 > 0.

(b) Deduce that the Gauss map of a surface M is conformal if and only ifM is either (i) a minimal surface without planar points or (ii) part of asphere.

4. (a) Show that total curvature is an isometric invariant for compact ori-entable surfaces. (Hint: Use Ex. 5 of Sec. 7.)(b) Extend (a) to the noncompact case, assuming K � 0 or K � 0.

5. (Total curvature of surfaces of revolution.) On a surface of revolution withprofile curve a, let Zab, as usual, be the zone bounded by the paral-lels through a(a) and a(b), a < b. Show that the total curvature of Zab is 2p(sinja - sinjb), where ja and jb are the slope angles of a at a and b—measured relative to the axis of revolution. (See Fig. 6.19.)

For a on an open interval, -• � A < u < B � •, the total curvature canbe treated as an improper integral, giving the (possibly infinite) result

provided both limits exist.

2p j jlim sin lim sina A

ab B

bÆ Æ

-( )

± -EG F 2

EVV

E J EJ VV1 2 1= = ( ) =( )

, .


6. (Continuation.)(a) Show that every surface of revolution with closed profile curve has totalcurvature zero, and every augmented surface of revolution with two inter-cepts has total curvature 4p.(b) Find the total curvatures of (i) paraboloid of revolution, (ii) catenoid,(iii) bugle surface.

7. Let F: M Æ N be a mapping of oriented surfaces. Prove:(a) F is area-preserving (Ex. 6 of Sec. 7) if and only if F is a diffeomor-phism with Jacobian JF = ±1.(b) If F is an isometry, then it is area-preserving, but not conversely.

8. Let M be a noncylindrical ruled surface whose rulings are entire straightlines; assume K < 0.

(a) Show that the total curvature of M is -2L(d), where d is a director curvewith ||d || = 1. (Hint: See Ex. 11 of Sec. 5.6.)(b) Find the total curvature of the saddle surface z = xy by this method(see Ex. 12).

9. (Gauss maps of some minimal surfaces.) In each case show that the Gaussmap covers the sphere omitting exactly n points.

(a) Catenoid: n = 2.(b) Helicoid: n = 2.(c) Scherk’s surface (Ex. 5 of Sec. 5.5): n = 4.What are the total curvatures of these surfaces?

10. Let E = x(R2) be Enneper’s minimal surface (Ex. 15 of Sec. 5.6). (Theformula for x is that of x1 in the next exercise.)

(a) Compute the unit normal

Uu v

u v u v=1

+ +- - -( )

12 2 1

2 22 2, , .


FIG. 6.19

(b) Express U in terms of polar coordinates on R2, and show that (i) theGauss map G ª U is one-to-one and (ii) G maps E onto the unit sphere Sminus the south pole (0, 0, -1).(c) Deduce that E has total curvature -4p.It is known that every complete nonflat minimal surface has total curva-ture -4pm for some 1 � m � •, and that only the catenoid and Enneper’ssurface have the extreme value -4p. (See R. Osserman, A Survey ofMinimal Surfaces, Dover, New York, 1986.)

11. (Computer continuation.)(a) On the domain -2.5 � u, v � 2.5, plot the surface given by

for at least six values of t from t = 0 (saddle surface) to t = 1 (Enneper’s surface E ). Animate this series of plots, observing the formation of the twocurves at which E cuts across itself.

(b) Plot E, on the domain given in (a), as viewed successively from alongthe x axis, y axis, and z axis.

12. Parametrize the saddle surface S using polar coordinates r, J (insteadof rectangular u, v). Then

(a) Find the total curvature of S by an explicit integration.(b) Express the unit normal U to S in terms of polar coordinates. Showthat the resulting Gauss map is one-to-one. Determine its image and useCorollary 8.6 to check the result in (a).

13. (Computer optional.) (a) Find a parametrization of the monkey saddleM by polar coordinates (see Ex. 19 of Sec. 5.4).

(b) Plot the region r � 1.(c) Find the total curvature of M.

6.9 Congruence of Surfaces

Two surfaces M and in R3 are congruent provided there is an isometry Fof R3 that carries M exactly onto . Thus congruent surfaces differ only intheir positions in R3. For example, the surfaces

are congruent under a 45° rotation about the z axis.

M z xy M zx y

: =-

and : =2 2

2

MM

x t u v u v u v tu

uvv

vu, , , , ,( ) = -( ) + - + - +ÊËÁ

ˆ¯

2 23

23

2

3 30 .


To simplify the exposition, we assume that the surfaces in this section areorientable as well as connected.

9.1 Theorem If F is a Euclidean isometry such that F(M) = , then therestriction of F to M is an isometry of surfaces,

Furthermore, if M and are suitably oriented, then F preserves shape oper-ators, that is,

for all tangent vectors v to M.

In short, congruent surfaces are isometric and have essentially the sameshape operators.

Proof. We know from Chapter 4, Section 5, that the restriction

is a smooth mapping. Also, the differential maps of F and F agree ontangent vectors v to M. In fact, such a v is the initial velocity of a curve athat lies in M; hence F(a) = F(a). Thus

It follows that F* preserves dot products of tangent vectors to M, since F*has this property for all pairs of tangent vectors. Furthermore, F: M Æis one-to-one (since F is) and onto (by hypothesis). Hence F is an isome-try of surfaces.

To show that F preserves shape operators, we arrange to have corre-sponding unit normals on M and . Let U be a unit normal to M. SinceF* preserves dot products and F carries M to , it follows that F*(U) isa unit vector field everywhere normal to . Thus F*(U ) is one of the twounit normals on , say,

(See Fig. 6.20.)If S and are the shape operators of M and derived from U and ,

we will show first that

F v F v* *S S( )( ) = ( )( )

UMS

F p F p* .U U( )( ) = ( )( )

MM

MM

M

F F* * .v F F v( ) = ( )¢ ( ) = ( )¢ ( ) = ( )a a0 0

F M M M= ÆF :

F S S F* *v v( )( ) = ( )( )

M

F M M M= ÆF : .

M

6.9 Congruence of Surfaces 315

for all tangent vectors to M.Again, let a be a curve in M with initial velocity v. Then F(a) is a curve

in with initial velocity F*(v). For U restricted to a and restricted to F(a), we have F(U) = (Fig. 6.20). Since F* preserves derivatives of vectorfields,

But v and S(v) are tangent to M, so F* can be replaced here by F*. ◆

Although congruent surfaces are isometric, it is not usually true that isometric surfaces are congruent. For example, the surfaces in Fig. 6.9 are all isometric to a plane rectangle, but evidently no two of them are congruent.

Our goal now is to prove the converse of the preceding theorem, namely,if M and are isometric and have essentially the same shape operators, thenthey are congruent. This is the analogue of the basic congruence theorem forcurves (Theorem 5.3 of Chapter 3). The condition M isometric to corre-sponds to the hypothesis that a and b are unit-speed curves defined on thesame interval, and “same shape operators” corresponds to

9.2 Theorem Let M and be oriented surfaces in R3. Let F: M Æbe an isometry of oriented surfaces in R3 that preserves shape operators, so

F S S F* *v v( )( ) = ( )( )

MM

k k t t= =, .

M

M

F v F FS U U S v( )( ) = - ¢( )( ) = - ¢( ) = ( )( )0 0 * .

UUM


FIG. 6.20

for all tangent vectors to M. Then M and are congruent; in fact, there isa Euclidean isometry F such that F|M = F.

(If it should happen that F*(S(v)) = - (F*(v)) for all tangent vectors, thenreversing the orientation of either M or will give the hypothesis as stated.)

Proof. There is only one candidate for F, which we describe as follows.Fix a point p0 in M and a tangent frame e1, e2 at p0. Let E3 and 3 be theunit normals to M and that give S and . By Theorem 2.3 of Chapter3, there is a unique Euclidean isometry F such that

These conditions imply that (F|M)* = F* at p0—a necessary condition for F|M = F.

Since M is connected, if p is an arbitrary point of M, there is a curvea in M from a(0) = p0 to p. It will suffice to show that F(a) = F(a), for thencertainly F(M) = and F|M = F.

There is no loss of generality in assuming that a lies in the domain of atangent frame field E1, E2 on M. If not, we could break up a into segmentsfor which this is true and in sequence repeat essentially the following proof.

We adjust this frame field, using the same constant rotation (andperhaps reflection) at each point, so that its value at p0 is the selected framee1, e2. This precaution is necessary in order that all choices of p in M willrelate properly to the Euclidean isometry F defined above.

Next, this frame field is transferred, by the isometry F, to a frame field 1, 2 on . Appending the selected unit normals gives an adapted frame

field E1, E2, E3 on a domain in M, and correspondingly 1, 2, 3 on .Preparations are completed by restricting these frame fields to the curves a and = F(a), respectively, as shown in Fig. 6.21.

The proof consists in applying Theorem 5.7 of Chapter 3 to the curvesa and . So we must verify the conditions (1) and (2) in that theorem.

Since the isometry F preserves velocities of curves and covariant deriv-atives of vector fields,

Being tangent to M, a¢ is orthogonal to E3; similarly for ¢ and 3. Thus ¢ • E3 = 0 = ¢ • 3. So we now have

(1)

Let wij and ij be the connection forms of the adapted frame fields {Ei}and { i}. By Lemma 5.3, F*( 12) = w12, and we computewE

w

¢ = ¢ ( )a a• • .E E ii i 1 3� �

EaaEa

¢ = ¢( ) ( ) = ( )¢ = ¢ =( )a a a a• * • * • • , .E F F E F E E kk k k k 1 2

a

a

MEEEMEE

M

F e e F e e F p p* * , * * , * .1 1 2 2 3 0 3 0( ) = ( ) ( ) = ( ) ( )( ) = ( )( )F F E E F

SME

MS

M

6.9 Congruence of Surfaces 317

Since F preserves shape operators,

The same result holds with 1 replaced by 2, so E2¢ • E3 = 2¢ • 3. By skew-symmetry in i and j, we conclude that

(2)

Equations (1) and (2) above are those of Theorem 5.7 in Chapter 3 (withb replaced by ). Furthermore, by construction, the Euclidean isometryproduced by the theorem is the isometry F defined above. ◆

This theorem provides a formal proof that the shape operators of a surfacein R3 do, in fact, completely describe its shape.

Exercises

1. A surface M Ã R3 is rigid provided every surface in R3 isometric to M iscongruent to M (thus M has only one possible shape in R3).

Deduce from Liebmann’s theorem that spheres are rigid, that is, if M Ã R3

is isometric to a “round” sphere S: || p - p0|| = r, then M is a round sphere.

2. If a,b: I Æ R3 are unit-speed curves with the same curvature k > 0 andtorsion t, show that their tangent surfaces are congruent. (Compare Ex. 5 inSec. 4, where only k is the same.)

3. Let M and N be congruent surfaces in R3, with F a Euclidean isometrysuch that F(M) = N. Prove that the isometry F|M preserves Gaussian andmean curvature, principal curvatures, principal directions, umbilics, asymp-

a

¢ = ¢ £ £( )E E E E i ji j i j• • , .1 3

EE

¢ = — = -— = ¢( )= ¢( )( ) = ¢( ) = — = ¢

¢ ¢

¢

E E E E E E S EF S F E S E E E E E

1 3 1 3 3 1 1

1 1 1 3 1 3

• • • •

* • * • • • .a a

a

aa a

¢ = — = ¢( ) = ( ) ¢( )= ¢( )( ) = ¢( ) = — = ¢

¢

¢

E E E E FF E E E E

1 2 1 2 12 12

12 12 1 2 1 2

• • *

* • • .a

a

w a w aw a w a


FIG. 6.21

totic and principal curves, and geodesics. Which of these are preserved byarbitrary isometries F: M Æ N?

4. For the sphere S Ã R3, show:(a) If F: S Æ S is an isometry, there is a Euclidean isometry F such thatF = F|S.(b) Same as (a), with F : S Æ S replaced by F : S Æ M Ã R3.

5. In each of the following cases, show that the surfaces are congruent byfinding an explicit Euclidean isometry F = TC that carries one surface to theother.

(a) z = xy and y = xz.(b) x2 - y2 - z2 = 1 and 2yz - x2 = 1.

6. If M is a surface in R3, a Euclidean isometry F such that F(M) = M iscalled a Euclidean symmetry of M. Show that

(a) The set of all Euclidean symmetries of M forms a subgroup S(M) ofthe group E (3) of all isometries of R3 (Ex. 7 of Sec. 3.1). S(M) is calledthe Euclidean symmetry group of M.(b) The Euclidean symmetric groups of congruent surfaces are isomorphic.

7. (a) Show that every Euclidean symmetry of the saddle surface

M: z = xy

is an orthogonal transformation C.(b) By considering the effect of C on the asymptotic unit vectors ±Ux and±Uy of M at 0, show that M has exactly eight Euclidean symmetries andfind their matrices.

8. Find all the Euclidean symmetries of the ellipsoid

where a > b > c.

(Hint: For the Gaussian curvature of the ellipsoid, see Example 5.2 of Ch. 5.)

6.10 Summary

The geometrical study of a surface M in R3 separates into three distinct categories:

(1) The intrinsic geometry of M.(2) The shape of M in R3.(3) The Euclidean geometry of R3.

xa

yb

zc

2

2

2

2

2

21+ + = ,

6.10 Summary 319

We saw in Chapters 2 and 3 that the geometry of R3 is based on the dotproduct and consists of those concepts preserved by the isometries of R3. Wehave now found that the geometry of M is also based on the dot product—applied only to vectors tangent to M—and that it consists of those featurespreserved by the isometries of M.

The shape of M in R3 is the link between these two geometries. Forexample, Gaussian curvature K is a crucial invariant of the intrinsic geome-try of M and (as Theorem 9.2 shows) the shape operator S dominates cate-gory (2). Thus the equation

K = det S

shows that the geometries (1) and (3) can be harmonized only by means ofrestrictions on (2). Stated bluntly: Only certain shapes in R3 are possible for asurface M with prescribed Gaussian curvature. A strong result of this type isLiebmann’s theorem, which asserts that a compact surface in R3 with K con-stant has only one possible shape—spherical.


▼

▲Chapter 7

Riemannian Geometry

321

In studying the geometry of a surface in R3 we found that some of its mostimportant geometric properties belong to the surface itself and not to thesurrounding Euclidean space. Gaussian curvature is a prime example;although defined in terms of shape operators, it beongs to this intrinsicgeometry since it passes the test of isometric invariance. In the 1850sRiemann drew the correct conclusion: There must exist a geometrical theoryof surfaces completely independent of R3, a geometry built from the startsolely of isometric invariants. In this chapter, we describe the fundamentalsof the resulting theory, concentrating on its dominant features: Gaussian cur-vature and geodesics. Our constant guides will be the two special cases thatled to its discovery: the intrinsic geometry of surfaces in R3 and Euclideangeometry—particularly that of the plane R2.

7.1 Geometric Surfaces

Evidence from earlier work on the intrinsic geometry of surfaces in R3 sug-gests that what is needed to do geometry on a surface is a dot product ontangent vectors. But to escape from confinement in R3, we must begin withan abstract surface M (Section 8 of Chapter 4). Since M need not be in R3

there is no dot product, and hence no geometry. However, the dot product isjust one instance of the general notion of inner product, and Riemann’s ideawas to replace the dot product by an arbitrary inner product on each tangentplane of M.

1.1 Definition An inner product on a real vector space V is a functionthat assigns to each pair of vectors v, w in V a number ·v, wÒ with the following properties.

(1) Bilinearity:

(2) Symmetry: ·v, wÒ = ·w, vÒ.(3) Positive Definiteness:

On the vector space R2 the dot product v • w = v1w1 + v2w2 is, of course, aninner product, but there are infinitely many others; for instance,

·v, wÒ = 3v1w1 + 2v2w2.

(See Exercise 4.)Basic features of the dot product remain valid for arbitrary inner products.

The length of a vector v is ||v|| = , and vectors are orthogonal if·v, wÒ = 0. The Schwarz inequality

allows the angle 0 £ J £ p between vectors to be defined by

Replacing surfaces in R3 by abstract surfaces and dot products by arbitraryinner products yields the following fundamental result.

1.2 Definition A geometric surface is an abstract surface M furnishedwith an inner product · , Ò on each of its tangent planes. These inner prod-ucts are required to vary smoothly in the sense that if V and W are differen-tiable vector fields on M, then ·V,W Ò is a differentiable real-valued functionon M.

We emphasize that each tangent plane Tp(M) of M has its own innerproduct, and these are unrelated except for the differentiability requirement—an obvious necessity for a theory founded on calculus. Here ·V,W Ò has itsusual pointwise meaning as the function assigning to each point p the number·V(p),W(p)Ò.

The geometric structure provided by this collection of inner products canbe described as a metric tensor g on M, that is, a function on all ordered pairsof tangent vectors v, w at points p of M such that

gp pv w v w, ,( ) = .

cos .J =v wv w

,

v w v w, £

· , Òv v

v v v v v, and , if and only if� 0 0 0; .= =

a a a a

b b b b

1 1 2 2 1 1 2 2

1 1 2 2 1 1 2 2

v v w v w v w

v w w v w v w

+ = +

+ = +

, , , ,

, , , .

322 7. Riemannian Geometry

Thus g is analogous to a differential 2-form on M (Definition 4.1 of Chapter4). But forms are skew-symmetric, w (w, v) = -w(v, w), while the metric tensoris symmetric, g(w, v) = g(v, w). (Informally, metric tensor is often shortenedto just metric.)

Definition 1.2 can be summarized as

We emphasize that the same surface furnished with two different metrictensors constitutes two different geometric surfaces.

The Euclidean plane R2, with its usual dot product, is the best-known geo-metric surface. Simple though it may be, R2 is the basic testing ground forthe geometry of surfaces. Of course, a surface M in R3 is a geometric surface,with the dot product of R3 applied, as usual, to tangent vectors on M. Thisgives the so-called induced metric on M, and unless some other geometricstructure is explicitly mentioned, it is always assumed that a surface in R3 hasthe induced metric.

1.3 Remark Construction methods.(1) Conformal Change. A simple way to get new geometric structures is to

distort old ones. For example, if h > 0 is a differentiable function on a regionin the Euclidean plane R2, define an inner product at each point p by

The resulting geometric surface M is said to be conformal with ruler functionh. In fact, the inner product of M gives the same angle measurements as thedot product on the Euclidean plane.

We call h > 0 a ruler function since larger values of h give smaller valuesfor the length of vectors (see Section 2). Examples will soon show that unlessh is quite special, the surface M has properties quite different from the Euclid-ean plane. In fact, locally every geometric surface can be so expressed.

(2) Pullback. A metric tensor g can be pulled back by a suitable mappingin the same way as differential forms are. Here is an important application.Let F: M Æ N be a regular mapping from an abstract surface M to a geo-metric surface N with metric tensor g = · , Ò. The pullback F*(g) of g to M isgiven in terms of inner products as

By definition, each tangent map F* is a linear isomorphism (linear, one-to-one, and onto), so it is easy to check that this definition gives an inner product

v w v w, * , *M NF F= ( ) ( ) .

v wv w

p, = ( )

•.

h2

surface metric tensor geometric surface+ = .

7.1 Geometric Surfaces 323

on each tangent space Tp(M). (The differentiability condition follows fromthe fact that F is a diffeomorphism on small enough neighborhoods.) ThusF*(g) is a metric tensor on M—in fact, it is the unique one that makes F alocal isometry.

(3) Coordinate description. Let x be a coordinate patch in an abstractsurface M (without geometry). If M were geometric, the functions

would be defined as before. But reversing this logic, if suitable functions E,F, G are given, then these equations determine a unique metric tensor · , Ò onthe image of x (see Exercise 4).

Many examples of these methods appear later on.From the modest beginning in Definition 1.2, a geometric theory can be

built that vastly enlarges that of Chapters 5 and 6. Definitions and theoremsfrom before that are clearly intrinsic in character will be used without furtherdiscussion. In particular, an isometry F: M Æ N of geometric surfaces is stilldefined by Definition 4.2 of Chapter 6, and the geometry of a geometricsurface M consists by definition of its isometric invariants.

Now that the calculus of R3 is gone, frame field computations on a surfaceitself become more important. A frame field on an arbitrary geometricsurface M consists, as usual, of two orthogonal unit vector fields E1, E2

defined on some open set in M. The orthonormality conditions

are now expressed, of course, in terms of the metric tensor of M. Two waysto construct frame fields are given in Exercises 7 and 9 of Section 2.

As before, dual 1-forms q1, q2 are uniquely determined by qi(Ej) = dij, andby Lemma 5.1 of Chapter 6, the connection form w12 = -w21 is uniquely char-acterized by the first structural equations

We emphasize that these forms q1, q2, w12 are not invariants of the surfaceM; a different choice of frame field , will produce different forms 1,

2, . To obtain invariant results we will need to know how two such setsof forms are related.

On a small enough neighborhood of a point p Œ M, careful use of theinverse functions cos-1 and sin-1 will yield a differential angle function j suchthat

E E E1 1 2= +cos sin .j j

w12qqE 2E1

d dq w q q w q1 12 2 2 21 1= Ÿ = Ÿ, .

E E E E E E1 1 1 2 2 21 0 1, , , , ,= = =

E F Gu u u v v v= = =x x x x x x, , , , ,


As Fig. 7.1 indicates, there are now two choices for E2. Either

so E1, E2 and , have the same orientation, or

and they have opposite orientation.In working in an oriented region, it is natural to use positively oriented

frame fields. Evidently any two such have the same orientation.

1.4 Lemma Let E1, E2 and , be frame fields on the same region inM. If these frame fields have

(1) the same orientation, then

(2) opposite orientation, then

Proof. We prove only (1), since (2) follows by changes of sign. By thebasis formulas in Lemma 2.1 of Chapter 6, the equations

yield

(*)

Taking the exterior derivative of the first of these gives

q j q j q q j q j q1 1 2 2 1 2= - = +cos sin sin cos .,

E E E

E E E

1 1 2

2 1 2

= +

= - +

cos sin

sin cos

j j

j j

,

w w j q q q q12 12 1 2 1 2= - +( ) Ÿ = - Ÿd , and .

w w j q q q q12 12 1 2 1 2= + Ÿ = Ÿd , and ;

E 2E1

E E E2 1 2= -sin cosj j ,

E 2E1

E E E2 1 2= - +sin cosj j ,


FIG. 7.1

Now substitute the first structural equations for , to obtain

In the same way, we get

Because the form w12 = -w21 uniquely satisfies the first structural equa-tions, we conclude from the last two equations that = w12 + dj, asrequired. Then direct computation of q1 Ÿ q2 using (*) shows that this 2-form equals 1 Ÿ 2. ◆

In this chapter, as earlier, the restriction to low dimensions is not essential.A surface is the 2-dimensional case of the general notion of manifold(Chapter 4, Section 8). A manifold of arbitrary dimension furnished with ametric tensor is called a Riemannian manifold. Euclidean geometry, as dis-cussed in Chapter 3, is the special case of Riemannian geometry producedon Rn by the usual dot product.

Thus a geometric surface is the same thing as a 2-dimensional Riemannianmanifold, and the subject of this chapter is 2-dimensional Riemannian geometry.†

Exercises

1. In a conformal geometric surface with ruler function h (Remark 1.3(1)),show that:

(a) The speed of a curve a = (a1, a2) is (b) hU1, hU2 is a frame field with dual 1-forms (c) The area forms are (d) The identity map from the Euclidean plane to this surface is a confor-mal map with scale factor (Def. 4.7 of Ch. 6). (Using h rather thanl as the descriptive function leads to simpler formulas.)

2. The Poincaré half-plane is the upper half-plane v > 0 in R2 with metrictensor . If a is the curve a(t) = (rcos t, r sin t), 0 < t < p,with constant r > 0,

v w v w p, • /= ( )v 2

l = 1/h

± Ÿdu dv h/ 2

du h dv h/ , / .¢ + ¢ ( )a a a1

22

2 / .h

qq

w12

d dq w j q2 12 1= - -( ) Ÿ .

d d

d

q w j j q j q

w j q

1 12 1 2

12 2

= -( ) Ÿ +( )= -( ) Ÿ

sin cos

.

dq2dq1

d d d d dq j j q j q j j q j q1 1 1 2 2= - Ÿ + - Ÿ -sin cos cos sin .


† We would prefer to call a geometric surface a Riemann(ian) surface, but this term has a firmlyestablished and distinctly different meaning.

(a) Show that the speed of a is csc t.(b) Deduce that although the Euclidean length of a is pr, its Poincarélength is infinite.(c) Find the area of the region between a and the u axis.

For a surface M in R3 oriented by a unit normal U, the rotation operatorJ was defined by J(v) = U ¥ v. Thus if M is an arbitrary geometric surface,we need a new definition for J so that the results at the end of Section 8,Chapter 6, will remain valid on M.

3. (Rotation operator.) Let M be a geometric surface oriented by an areaform dM. Prove:

(a) On each tangent space to M there exists a unique “rotation by +90°,”that is, a linear operator J: Tp(M) Æ Tp(M) such that

(Hint: If J is the linear operator such that J(e1) = e2, J(e2) = -e1 holds forone positively oriented frame, show that these relations hold for every suchframe.)(b) J is differentiable (that is, V differentiable implies J(V) differentiable),skew-symmetric (that is, ·J(v), wÒ = -·v, J(w)Ò), and J 2 = -I, that is,J(J(v)) = -v for all v.(c) If M is oriented instead by -dM, then its rotation operator is -J.(d) If M is a surface in R3 oriented by a unit normal U and dM is the areaform such that dM(v, w) = U • v ¥ w, then for J as in (a), J(z) = U ¥ zfor all z.

4. (Coordinate definition of a metric.)(a) If a, b, and c are numbers such that a > 0, c > 0, ac - b2 > 0, then theformula

defines an inner product on R2. (Hint: .)

(b) Let x: D Æ M be a coordinate patch in an abstract surface M (withoutgeometry). Given differentiable functions E, F, and G on D such that

prove that there is a unique metric tensor · , Ò on the image of x such that

E F Gu u u v v v= = =x x x x x x, , , , , .

E G EG F> > >0 0 2, ,

av cv1 2

2

0±( ) ≥

x xu v av w b v w v w cv w, = + +( ) +1 1 1 2 2 1 2 2

J J dM Jv v v v v v v( ) = ( ) = ( )( ) > π( ), , , , if0 0 0 .


5. (Line-element.)(a) For an arbitrary inner product, prove the polarization identity

(b) Deduce that the metric tensor of a geometric surface M is completelydetermined by the function q whose value on each tangent vector v is ||v||2.

Classically, q is called the line-element, or first fundamental form, of M andis expressed in coordinates as

(Here du dv, for example, is ordinary multiplication, not wedge product.)(c) Account for the unusual notation ds2 by finding the relevant formulafor the speed ds/dt of a curve a(t) = x(u(t), v(t)) (compare Ex. 5 of Sec.5.4).

6. If F: M Æ N is a regular mapping of oriented geometric surfaces, showthat the following are equivalent:

(a) F is orientation-preserving and conformal (Def. 4.7 of Ch. 6).(b) F preserves rotation operators; that is, for all tangent vectors v,

(c) F preserves oriented angles; that is, if j is an oriented angle from v tow, then it is also an oriented angle from F*(v) to F*(w).

7. Prove that a regular mapping F = ( f,g) from a region U Ã R2 into R2 isconformal and orientation-preserving if and only if the Cauchy-Riemannequations fu = gv, fv = -gu hold.

If R2 is considered as the complex plane, with z = (u, v) = u + iv, these equations are necessary and sufficient for z Æ F(z) to be a complex analyticfunction. Show also that the scale factor of the resulting conformal map is

8. (Continuation.) If the origin is deleted from R2, show that the mapping F in (2) of Example 7.3 of Chapter 1 is conformal and orientation-preserving. What is the complex function in this case?

9. Let M be a region U Ã R2 furnished with a conformal metric given byruler function h. If F is a Euclidean isometry U Æ U that preserves h (thatis, h(F ) = h), show that F is an isometry M Æ M.

dF dz/ .

F J J FM N* *v v( )( ) = ( )( ).

q ds E du F du dv G dv= = + +2 2 22 .

v w v w v w, = + - -( )14

2 2 .


7.2 Gaussian Curvature

For geometric surfaces we need a new definition of Gaussian curvature. Thedefinition K = det S for surfaces in R3 is meaningless now, since it uses shapeoperators. But this original definition made K an isometric invariant, so it isreasonable to look at the proof of the theorema egregium—specifically toCorollary 2.3 of Chapter 6—to find a satisfactory generalization.

2.1 Theorem On a geometric surface M there is a unique real-valuedfunction K such that for every frame field on M the second structural equation

holds. K is called the Gaussian curvature of M.

Proof. For a frame field E1, E2, it follows from the basis formulas inLemma 2.1 of Chapter 6 that there is a unique function K such that

dw12 = -Kq1 Ÿ q2.

But a priori this function depends on the particular frame field: Anotherframe , might have a different function for which

Evidently we must show that where the domains of the frame fields overlap,K = . Such domains cover all of M, so we will then have a single func-tion K with the required property.

Suppose first that the two frame fields have the same orientation. ByLemma 1.4, = w12 + dj. Hence d = dw12, since d 2 = 0. But then

Since

we conclude that = K.When the orientations are opposite, we get d = -dw12, but still find= K, since

◆

As mentioned above, Corollary 2.3 of Chapter 6 shows that this general def-inition of Gaussian curvature agrees with the definition K = det S when Mis a surface in R3. The proof of isometric invariance presented there is entirely

q q q q1 2 1 2Ÿ = - Ÿ .

Kw12

K

q q q q1 2 1 2 0Ÿ = Ÿ π ,

K Kq q q q1 2 1 2Ÿ = Ÿ .

w12w12

K

d Kw q q12 1 2= - Ÿ .

KE 2E1

d Kw q q12 1 2= - Ÿ


intrinsic in character and thus holds for arbitrary geometric surfaces. Fur-thermore, the orthogonal curvature formula in Proposition 6.3 of Chapter 6was derived from dw12 = -Kq1 Ÿ q2, without reference to the shape operator;hence it is valid for any geometric surface.

To summarize: Geometrical investigations in terms of a frame field E1, E2

are expressed in terms of its structural equations

The first structural equations determine the connection form w12 = -w21 ofthe frame field, but the second structural equation defines the Gaussian curvature of the geometric surface—independent of choice of frame field.Chapter 6, Section 6 showed how w12 and K can be explicitly computed fromthese implicit definitions.

Gaussian curvature is the central property of a geometric surface M. Wewill see that it influences—often decisively—virtually every aspect of thegeometry of M.

Let us make certain that the new definition of curvature gives the correct result for the Euclidean plane R2. Its natural frame field U1 = (1, 0), U2 = (0, 1) has dual 1-forms q1 = du, q2 = dv. Then dq1 = dq2 = 0,so the identically zero 1-form w12 = 0 satisfies the first structural equationsand hence is the connection form of U1, U2. Obviously, dw12 = 0, so by thepreceding theorem, K = 0. It can be no surprise that the Euclidean plane isflat, for R2 is isometric to, say, the xy plane in R3, which has K = 0 since itsshape operators all vanish.

2.2 Example A flat torus. Let T be a torus of revolution considered asan abstract surface, without geometry. If x is its usual parametrization(Example 2.5 of Chapter 4), then the definitions

determine a unique inner product on each tangent plane of T. The resultinggeometric surface T0 is certainly different from the torus of revolution: it isflat. To see this, recall that

Thus x* carries frames to frames, and consequently x is a local isometry of theEuclidean plane R2 onto T0—and local isometries preserve curvature. ◆

In effect, T0 is constructed by gluing opposite sides of a Euclidean rectan-gle, as specified by the map x. This metric on the flat torus is not the induced

x x x x* and *U Uu v1 2( ) = ( ) = .

x x x x x xu u u v v v, , , , ,= = =1 0 1

d d

d K

q w q q w q

w q q

1 12 2 2 21 1

12 1 2

= Ÿ = Ÿ

= - Ÿ

, ,

.


metric derived, as usual, from the dot product of R3. If it were, then since T0

is compact, by Theorem 3.5 of Chapter 6 it would have to have positive cur-vature somewhere. Thus T0 can never be found in R3.

This shows that the class of all geometric surfaces is larger than that ofsurfaces in Euclidean 3-space—in fact, it is immensely larger. In the courseof this chapter and the next, we hope to persuade the reader that geometricsurfaces are the natural objects to study and that surfaces in R3—howevervisually appealing—are only an interesting special case.

We now find a formula for the Gaussian curvature of the conformal geo-metric surfaces described in Remark 1.3.

2.3 Corollary For the plane R2 with metric tensor ,the Gaussian curvature is

Proof. Let M denote the plane with its new metric. Then the identitymap R2 Æ M is a patch with and F = 0. Thus the resultfollows directly from the orthogonal curvature formula in Proposition 6.3 of Chapter 6. (Since the patch is conformal, we could also get K as h2 D(log h) from Exercise 2 of Section 6.6) ◆

2.4 Example (1) The stereographic sphere. It was shown in Example 5.5of Chapter 4 that stereographic projection P is a diffeomorphism of the punctured sphere S 0 onto the Euclidean plane R2. Now consider S 0 as merely an abstract surface, and assign it the pullback metric tensor ofRemark 1.3(2) that makes P an isometry. If S 0 appears round in Fig. 7.2, itis only because we look at it with Euclidean eyes—intrinsically, this S 0 is asflat as the plane.

(2) The stereographic plane. Now reverse the process in (1). Consider S 0

with its usual geometric structure as a surface in R3 and ignore the geometryof R2.

E G h= = 1 2/

K h h h h huu vv u v= +( ) - +( )2 2 .

v w v w p, • /= ( )h2


FIG. 7.2

The inverse P-1: R2 Æ S 0 of stereographic projection is a diffeomorphismsince P is. Hence the pullback operation makes R2 a geometric surface—thestereographic plane—that is isometric to S 0 Ã R3 and hence has curvature K = 1.

We examine this stereographic plane more closely. If v and w are tangentvectors to R2 at q = P(p), let and be the unique tangent vectors to S 0 thatP* carries to v and w, respectively. By Exercise 4.14 of Chapter 6, we knowthat in terms of Euclidean dot products,

But (P-1)* = (P*)-1 carries v and w back to and , so for the pulled backinner product on R2 we find

It follows immediately that the new metric on the stereographic plane is con-formal as in Remark 1.3, with ruler function

To visualize this unusual plane, it helps to imagine that rulers get longer asthey move farther from the origin. Since P is now an isometry, the intrinsicdistance from p to q in Fig. 7.3 is exactly the same as the distance from p* toq*. Also, circles u2 + v2 = r2 with r very large actually have very small stereo-graphic radii, since they correspond under P to small circles about the(missing) north pole of S 0.

h u vu v

, .( ) = ++

14

2 2

v w v w v wq

v w, * *= ( ) ( ) ( ) ( ) = = +ÊËÁ

ˆ¯

- -

-

P P1 1

2 2

14

• ˜ • ˜ • .

wv

v w v wq

v w• ˜ • ˜ ˜ • ˜ .= ( ) ( ) = +ÊËÁ

ˆ¯

P P* * 14

2 2

wv


FIG. 7.3

2.5 Example The hyperbolic plane. Let us experiment with a change ofsign in the stereographic inner product above, setting

Since h > 0 is necessary, this hyperbolic inner product is usedonly on the open disk u2 + v2 < 4 of radius 2 in R2. The resulting geometricsurface is the Poincaré disk model of the hyperbolic plane H. It follows easilyfrom Corollary 2.3 that the hyperbolic plane has constant Gaussian curvatureK = -1. ◆

As a point (u, v) approaches the rim of H, that is, the circle u2 + v2 = 4 (notpart of H!), h(u, v) approaches zero. In the language used above, rulers shrinkas they approach the rim, so H is a good deal bigger than Euclidean intu-ition may suggest.

To verify this, let us find the arc length function s(t) of the Euclidean ray

which runs, at angle J, from the center of H out to the rim. On g, the rulerfunction becomes 1 - t2/4. Since g ¢ = (cosJ, sinJ),

Then

Thus, as t approaches 2, s(t) approaches •. This “short” segment has infinitehyperbolic length. (Although distance in H is quite non-Euclidean, recall thatby conformality, angle measurement is the same.)

Further properties of the hyperbolic plane will be developed as we goalong. We will see that it—and not the bugle surface (Example 7.6 in Chapter5)—is the true analogue of the sphere for constant negative curvature.

Any regular mapping F: M Æ N can be used to pull a metric tensor on Nback to M, but unless F is a diffeomorphism, it is usually not possible to pushforward a metric from M to N. The trouble is that when F carries points p1 π p2 of M to the same point of N, the inner products at these points maytransfer differently to F(p1) = F(p2). This difficulty can be eliminated by impos-ing a consistency condition, as follows.

2.6 Proposition Let F be a regular mapping of a geometric surface Monto a surface N without geometry. Suppose that whenever F(p1) = F(p2) there

s tdu

u

t tt

t( ) =

- ( )= =

+-Ú -

1 22

2222

0

1tanh log .

dsdt

th t

= ¢( ) = ( ) =- ( )

gg1 1

1 22 .

g J Jt t t t( ) = ( ) <( )cos sin, ,0 2�

v w v w, • /= h2

h u vu v

, .( ) = -+

14

2 2


is an isometry G12 from a neighborhood of p1 to a neighborhood of p2 suchthat

Then there is a unique metric tensor on N that makes F a local isometry.

Proof. If F is to be a local isometry, there is no choice in defining theinner product of tangent vectors v, w at q in N. F is regular, so at any pointp1 in M such that F(p1) = q there are unique vectors v1, w1 such that

Hence we must define ·v, wÒN to be ·v1, w1ÒM.For this to be a valid definition, it must not depend on the choice of p1.

Explicitly, if p2 is another point such that F(p2) = q and if v2, w2 are theunique vectors there such that

then we require

(*)

To prove this, let G = G12 be an isometry as in the statement of the propo-sition. Since FG = F, the chain rule gives F*G* = F*. Consequently, G*carries the vectors v1, w1 to the vectors at p2 that F* carries to v, w, namely,to v2, w2. Since G is an isometry, (*) holds. ◆

2.7 Example The projective plane. Example 8.2 of Chapter 4 defined theprojective plane P as an abstract surface by identifying antipodal points of aunit sphere S Ã R3. Now we give P a geometric structure. Recall that the pro-jection F: S Æ P is related to the antipodal map A(p) = -p by FA = F. SinceA is an isometry, the preceding result applies, with all maps G12 being A. Theresulting local isometry S Æ P shows that like the sphere, P has constantpositive curvature K = 1 and all its geodesics are closed.

The same scheme, applied when S has radius r, gives the projective planeof radius r, with curvature ◆

Like the flat torus, P cannot be found in R3. For the flat torus this failurecould be blamed on its geometry, since tori of revolution abound in R3. Butfor P the obstruction is topological, because (by Theorem 7.10 of Chapter 4)compact surfaces in R3 must be orientable—and P is not.

K r= -1 2/ .

v w v w2 2 1 1, ,M M

= .

F F* , * ,v v w w2 2( ) = ( ) =

F F* , *v v w w1 1( ) = ( ) = .

FG F G12 12 1 2= ( ) =, p p .


Now that we have a geometry for surfaces that may not occur in R3, anatural place to look for them is in higher-dimensional Euclidean spaces. Inthe simplest case, let x: D Æ Rn, n ≥ 3, be a regular map, so x(D) is at leastan immersed surface. Then pullback of the dot product metric of Rn makesD a geometric surface, with

We think of D as a geometric description of x(D), as in the n = 3 case.

2.8 Example Tangent surfaces. For any n ≥ 3, let b be a unit-speed curvein Rn with k = ||T ¢|| > 0. The tangent surface is given as before by the formulax(u, v) = b(u) + vT(u), with v π 0. Then xu = T + vT ¢ and xv = T, so

Since EG - F 2 = v2k2(u) > 0, x is regular. The resulting immersed surface isflat, just as in the R3 case. This follows from the general curvature formula inExercise 9, but it is clear without computation since we know that K derivessolely from E, F, and G, and these are given by the same expressions as inthe 3-dimensional case (Exercise 13 in Section 5.4). ◆

Exercises

1. Show that the Poincaré half-plane (Ex. 2 of Sec. 1) has constant nega-tive Gaussian curvature K = -1.

2. For the conformal structure (Rmk. 1.3) on the entire plane with rulerfunction h = sech(uv), find the dual forms and connection forms of the framefield hU1, hU2 and derive the Gaussian curvature K. Check by finding K fromCorollary 2.3.

3. Find the area of the disk u2 + v2 � a2 in the hyperbolic plane. (Hint: Usepolar coordinates x(r, v) = (rcosv, r sinv).) What is the area of the entirehyperbolic plane?

4. The hyperbolic plane H(r) of pseudo-radius r is the disk u2 + v2 < 4r2 withconformal metric given by the ruler function

Thus the hyperbolic plane of Example 2.5 is H(1). Show that H(r) has con-stant Gaussian curvature K r= -1 2/ .

h u vu v

r, .( ) = -

+1

4

2 2

2

E v u F G= + ( ) = =1 1 12 2k , , .

v w x v x v, * *= ( ) ( )• .


5. In Example 2.2 suppose the flat torus T0 is constructed from a torus ofrevolution T with radii R > r > 0. Find the area of T0. How does this comparewith the area of T with its usual geometry as a surface in R3?

6. (a) Show that the geometric surface in Exercise 2 is isometric to a heli-coid. (Hint: Use x(u, v) = (f(u) cosv, f(u) sinv, v) for a suitable function f(u).)(b) Check that the two surfaces have the same Gaussian curvature at cor-responding points.

7. A scale change of M stretches every dimension of M by a constant factorc > 0. Formally, if M is a geometric surface with inner product · , Ò, let bethe same abstract surface with inner product

We say that is M scaled by c > 0. Show that:

(a) ||v||- = c||v|| for all v, but angles between vectors are the same.(b) For any curve segment a, (a) = cL(a).

(c) Frame fields E1, E2 on M and on have dual 1-forms i = cqi, but their connection forms are equal.

(d) A region R has M area A if and only if it has area c2A.(e)

Taking c > 1 for definiteness, the scale change expands every dimension of Mby the factor c—which produces smaller curvature (compare a sphere ofradius 2 with a unit sphere). It will follow from Lemma 4.5 that M and have the same geodesics—though speeds differ.

8. Prove:(a) The sphere S(r) Ã R3 of radius r is isometric to the unit sphere S scaledby r.(b) The hyperbolic plane H(r) of Exercise 4 preceding is isometric to H = H(1) scaled by r.

Because the geometric effects of scale change are so simple, it usually sufficesto work with standard models such as S and H.

9. (Classical tensor formula for Gaussian curvature.) For an arbitrary patchx, the associated frame field is

where as usual, .W EG F= - 2

EE

EW E

E Fuv u1 2

1= = -( )x

x x, ,

M

K K c= / .2

Mq

ME c2 /E c1 / ,

L

M

v w v w, ,- = c2 .

M


(E2 is found by subtracting from xv its E1 component and reducing theresult to unit length—the so-called Gram-Schmidt process.)

(a) Check that E1, E2 is orthonormal.(b) Express the dual 1-forms q1, q2 in terms of du and dv.(c) Find the functions P and Q such that w12 = Pdu + Qdv.(d) Deduce the curvature formula

This formula is usually stated in terms of Christoffel symbols Gkij, a set of

functions derived from E, F, and G.(e) For an orthogonal patch, show that (d) reduces to Proposition 6.3 ofChapter 6.(f) Test (d) for the saddle surface in Example 4.3(2) of Chapter 5.

10. (Continuation.) If the parameter curves of a patch all have unit speed,show that Gaussian curvature is given by , where 0 < J < p isthe coordinate angle.

11. (a) Show that the mapping x: R2 Æ R4 given by

is conformal, hence regular. (In fact, x parametrizes a surface in R4.)(b) Find its Gaussian curvature. (Hint: Ex. 2 of Sec. 6.6.)

12. (Double saddle.) The image of the patch is a surface in R4. Find:

(a) its Gaussian curvature;(b) the area of the region for which 0 � u, v � 1.

13. (Computer.)(a) Write the computer command based on Exercise 9 that, given the func-tions E, F, G, returns K.(b) Show how (a) can be used to find the curvature of x: D Æ Rn for arbitrary n.(c) Test (a) on the three preceding exercises and on Example 2.8.

7.3 Covariant Derivative

The covariant derivative — of R3 (Chapter 2, Section 5) is an essential partof Euclidean geometry. It is used, for example, to define the shape operator

x u v u v uv u v, , , , /( ) = -( )( )2 2 2

x u v u v u v u v u v, , , ,( ) = ( )cosh cos cosh sin sinh cos sinh sin

K uv= -J J/sin

KW u

FE EGEW v

EF FE EEEW

v u u u v=∂

∂-Ê

ËÁˆ¯ +

∂∂

- -ÊËÁ

ˆ¯

ÈÎÍ

˘˚

12

2.

7.3 Covariant Derivative 337

of a surface in R3 and, in modified form, to define the acceleration of a curvein R3 (Chapter 2, Section 2). In this section we show that each geometricsurface has its own notion of covariant derivative.

As in Euclidean space, a covariant derivative — on a geometric surface Massigns to each pair of vector fields V, W on M a new vector field —VW. Intu-itively, the value of —VW at a point p will be the rate of change of W in theV(p) direction. We must certainly require that — have the familiar linear andLeibnizian properties (1)–(4) in Corollary 5.4 of Chapter 2—of course, withthe Euclidean dot product replaced by the inner product of M.

Furthermore, since the connection form w12 of a frame field E1, E2 mea-sures the rate at which E1 is turning toward E2, we must also require

(*)

These conditions are enough to completely determine —.

3.1 Lemma Assume that there is a covariant derivative — on M with theusual linear and Leibnizian properties, and such that (*) holds for frame fieldsE1, E2. Then — obeys the connection equations

Furthermore, for a vector field W = f1E1 + f2E2,

We call this last equation the covariant derivative formula. Note that in thisformula, V [ f1] and V [ f2] only tell how W is changing relative to the framefield. The other two terms tell how the frame field itself is rotating, so thecombined formula makes —VW an absolute rate of change.

Proof. By orthonormal expansion,

The first summand vanishes since ·E1, E1Ò is constant. In fact, by a famil-iar Leibnizian argument,

Using (*) in the second summand gives —VE1 = w12(V)E2, as required. Theother connection equation is derived similarly.

Finally, the properties of — yield

0 21 1 1 1= [ ] = —V E E E EV, , .

— = — + —V V VE E E E E E E1 1 1 1 1 2 2, , .

— = [ ] + ( )( ) + [ ] + ( )( )VW V f f V E V f f V E1 2 21 1 2 1 12 2w w .

— = ( ) — = ( )V VE V E E V E1 12 2 2 21 1w w, .

w12 1 2V E EV( ) = — , .


Substituting the connection equations then gives the covariant derivativeformula. ◆

This lemma shows how to define the covariant derivative of M. The fol-lowing result (here stated for dimension 2) has been called the fundamentaltheorem of Riemannian geometry.

3.2 Theorem On each geometric surface M there exists a unique covari-ant derivative — with the linear and Leibnizian properties (1)–(4) in Corol-lary 5.4 of Chapter 2 and satisfying equation (*) for every frame field.

Proof. The preceding lemma shows that there is at most one such covari-ant derivative since it gives an explicit formula for —VW. The proof thatthere is at least one splits into two parts.

Local Definition. For a frame field E1, E2 on a region U in M, use thecovariant derivative formula in Lemma 3.1 as the definition of —VW. It isa routine exercise in calculus on a surface to verify that — has the requiredproperties. The linearity conditions (1) and (2) are simple, so we will provethe Leibnizian property (3), namely,

Write Y = g1E1 + g2E2. Then fY = fg1E1 + fg2E2, and by the covariantderivative formula,

The Leibnizian product rule V [ fgi] = fV[gi] + giV [ f ] yields

But the right side here is f —VY + V [ f ]Y, as required.To prove (*), use the covariant derivative formula with W = E1. Then

f1 = 1 and f2 = 0, so the formula collapses immediately to (*).

Consistency. For two different frame fields, do the local definitionsagree? If so, we have a single covariant derivative well-defined on all of M.So let derive from a frame field , on a domain . We must showUE 2E1—

— ( ) = [ ] + ( )( ) + [ ] + ( )( )+ [ ] + [ ]

V fY f V g g V E f V g g V E

g V f E g V f E

1 2 21 1 2 1 12 2

1 1 2 2

w w

.

— ( ) = [ ] + ( )( ) + [ ] + ( )( )V fY V fg fg V E V fg fg V E1 2 21 1 2 1 12 2w w .

— ( ) = [ ] + —V VfY V f Y f Y .

— = — +( ) = — ( ) + — ( )= [ ] + — + [ ] + —

V V V V

V V

W f E f E f E f E

V f E f E V f E f E

1 1 2 2 1 1 2 2

1 1 1 1 2 2 2 2 .


that —vW = VW on the intersection of U and . Because of the prop-erties of —, it will suffice to show

We use Lemma 1.4, assuming for simplicity that the two frame fields havethe same orientation. When —V is applied to the equation

the covariant derivative formula gives

By Lemma 1.4, = w12 + dj. Since dj(V) = V [j], substituting w12 = - dj into the preceding equation reduces it to

In the same way, —V = V follows from = -sinjE1 + cosjE2. ◆

3.3 Example The covariant derivative of R2. The natural frame field U1,U2 has w12 = 0. Thus, for a vector field W = f1U1 + f2U2, the covariant deriv-ative formula (Lemma 3.1) reduces to

This is just Lemma 5.2 of Chapter 2, applied on R2 instead of R3, so ourabstract definition of covariant derivative produces correct Euclidean results.

Note that the covariant derivative formula shows that (as in the Euclideancase) the value of the vector field —VW at a point p depends only on W andthe tangent vector V(p). Thus —vW is meaningful for an individual tangentvector.

The covariant derivative on a geometric surface M can be adapted to avector field Y along a curve a: I Æ M. (Recall that Y assigns a vector Y(t)in Ta(t)(M) for each t in I.) The covariant derivative Y¢ of Y ought to be —a ¢Y, but neither a ¢ nor Y is defined on an open set of M as required by thedefinition of —. The simplest solution is to define Y¢ by a frame field formulamodeled on the covariant derivative formula in Lemma 3.1.

So for a frame field E1, E2, write Y = f1E1 + f2E2, and then define

— = [ ] + [ ]VW V f U V f U1 1 2 2 .

E 2E 2—E 2

— = ( ) - +( )= ( ) = —

V

V

E V E E

V E E

1 12 1 2

12 2 1

w j j

w

sin cos

.

w12

w12

— = - [ ] + ( )( )+ [ ] + ( )( )

V E V V E

V V E

1 21 1

12 2

sin

cos .

j j w

j j w

E E E1 1 2= +cos sinj j ,

— = — — = —V V V VE E E E1 1 2 2, .

U—


(d)

Standard proofs show that this definition is independent of the choice offrame field and has the same linear and Leibnizian properties as in the Euclid-ean case.

The velocity a ¢ of a curve in M is a vector field on M, so we can take itscovariant derivative to get the acceleration a ≤ of a.

In the frequently occurring case of vector fields of constant length, thereis a more intuitive description of the covariant derivative along a curve—onethat relates more directly to other geometric invariants.

3.4 Lemma Let E1, E2 = J(E1) be a positively oriented frame field on M,where J is the rotation operator from Exercise 3 of Section 7.1. Let Y be avector field of constant length c > 0 along a curve a in M. If j is an anglefunction from E1 to Y, then

Again, j ¢ tells the rate at which Y is rotating relative to the frame field,and the connection term tells how the frame field is rotating along a, so theirsum gives an absolute rate for Y.

Proof. We write Then applying the covariantderivative formula gives

Multiplication by c completes the proof. ◆

A distinctive feature of Euclidean geometry is the ability to move a tangentvector v at one point to a parallel vector at any other point by simply keepingthe same natural coordinates for v. As we shall see, this phenomenon of“distant parallelism” rarely occurs on an arbitrary geometric surface.However, it is always possible to define parallelism of a vector field along acurve.

3.5 Definition A vector field V on a curve a in a geometric surface is parallel provided its covariant derivative vanishes: V¢ = 0.

¢ = - ¢ + ¢( )( ) + ¢ + ¢( )( )= ¢ + ¢( )( ) -( ) = ¢ + ¢( )( ) ( )

Y c E E

E E J Y c

/ sin cos

cos sin / .

j j w a j j w a

j w a j j j w a

21 1 12 2

12 2 1 12

Y c E E/ cos sin .= +j j1 2

¢ = ¢ + ¢( )( ) ( )Y J Yj w a12 .

¢ = ¢+ ¢( )( ) + ¢+ ¢( )( )Y f f E f f E1 2 21 1 2 1 12 2w a w a .


As in the Euclidean case, a parallel vector field has constant length, since||V ||2 = ·V, V Ò and ·V, V Ò¢ = 2·V ¢, V Ò = 0.

3.6 Lemma Let a be a curve in a geometric surface M, and let v be atangent vector at p = a (t0). Then there is a unique parallel vector field V ona such that V(t0) = v (see Fig. 7.4).

Proof. The vector field V must satisfy the conditions

The existence and uniqueness of V is assured by differential equationstheory, but the following explicit proof may be more illuminating.

We can suppose that a lies entirely in the domain of a positively ori-ented frame field E1, E2 = J(E1) on M, for otherwise a could be broken intosubsegments for which this is true. Since V must have constant length c = ||v||, we can write

where the angle function j = j (t) is to be determined.By the preceding lemma, V will be parallel, that is, V¢ = 0, if and only if

j¢ + w12(a ¢) = 0. Furthermore, V(t0) = v will hold if j (t0) is an angle fromE1(p) to v. There is exactly one function j satisfying these conditions, namely

◆

For a parallel vector field V on a, we say that the vector V(t) at each pointa(t) is gotten from v at p = a(t0) by parallel translation along a.

In Euclidean space, parallelism means keeping the same natural coordi-nates, so parallel translation is path-independent. But in a geometric surface,parallel translation from p to q along different paths usually gives different

j j w at t dut

t( ) = ( ) - ¢( )Ú0 12

0

.

V c E c E= +cos sinj j1 2,

¢( ) = ( ) =V t t V t0 0for all , and v.


FIG. 7.4

results. Equivalently, if a vector v at p is parallel-translated around a closedcurve, the result v* need not be v—a phenomenon called holonomy.

If a: [a,b] Æ M is a closed curve in the domain of a frame field, the formulaj¢ + w12(a ¢) = 0 in Lemma 3.6 shows that all parallel vector fields along a arerotated through the angle

We call this the holonomy angle ya of a. (Additive multiples of 2p can beignored in ya since they do not affect the determination of v*.)

3.7 Example Holonomy on a sphere S of radius r. Suppose the closedcurve a parametrizes a circle on S . There is no loss of generality in assum-ing that a is a circle of latitude, say the u-parameter curve

where x is the geographical parametrization of S . For the associated frame field , E2 = J(E1) of x, Example 6.2 of Chapter 6 shows that w12 = sin vdu. Since v has constant value v0 on a, the holonomy angle ya ofa is

Thus only on the equator v = 0 does a vector v return to itself after paralleltranslation around a. If a is in the northern hemisphere (v0 > 0), then j¢ < 0,so an initially north-pointing parallel vector rotates eastward as it traversesa (Fig. 7.5).

Up to this point, everything in this section belongs to the geometry of anarbitrary surface, so it may be well to look back at the case of a surface Min R3. There we now have two ways to take covariant derivatives: one fromthe intrinsic geometry of M as a geometric surface, the other the Euclideancovariant derivative of R3. The two are usually different, but there is a simplerelationship between them. In the following lemma, — denotes the covariantderivative of M as a geometric surface and is the Euclidean covariant derivative.

3.8 Lemma If V and W are tangent vector fields on a surface M in R3,then, as in Fig. 7.6,

(1) —VW is the component of VW tangent to M.(2) If S is the shape operator of M derived from a unit normal U, then

—

—

j p j pa

p

2 0 200

2

0( ) - ( ) = - = - = -Ú Úsin sin sin .v du v du v

E E1 = ¢a

a pu u v u( ) = ( )x , for ,0 0 2� �

j j wa

b a( ) - ( ) = -Ú 12.


Proof. Since (2) implies (1), we need only prove (2). Let E1, E2, E3 = Ube an adapted frame on M. Suppose first that W is the vector field E1. Bythe Euclidean connection equations (Theorem 7.2 of Chapter 2),

The connection equations for M (Lemma 3.1) give

Since E3 = U, the definition of wij in Section 7 of Chapter 2 gives

Substitution then gives (2) for W = E1. The same result holds for E2.In the general case, write W = f1E1 + f2E2. Then (2) follows, by a routine

computation, from the special cases above. ◆

Evidently we have been using the intrinsic covariant derivative of M Ã R3

all along, without giving it formal recognition.For a patch x in an arbitrary geometric surface M, we shall inevitably use

the notation xuu for the covariant derivative of the vector field xu along u-parameter curves—with corresponding meanings for xuv = xvu and xvv. Thuswhen M is a surface in R3 we will need a new notation, say uu, . . . for thecorresponding Euclidean covariant derivatives.

x

w13 3 1 3 3 3 1 3 1V E E E E E E E S V E UV V( ) = —( ) = - —( ) = ( )( )˜ • ˜ • • .

— = ( )V E V E1 12 2w .

˜ .— = ( ) = ( ) + ( )=

ÂV j jj

E V E V E V E1 11

3

12 2 13 3w w w

˜ • .— = — + ( )( )V VW W S V W U


FIG. 7.5 FIG. 7.6

Exercises

1. In the Poincaré half-plane,(a) show that the connection form of the frame field E1 = vU1, E2 = vU2 is

Then for each of the following curves, express velocity and acceleration interms of this frame field:(b) The curve a in Exercise 2 of Section 1. (Hint: a ¢ and a ≤ are collinear.)(c) The Euclidean straight line

where c and s are constants such that c2 + s2 = 1. Check results using theformula ·b ¢, b ¢Ò¢ = 2·b ¢, b≤Ò.

2. Let V be a parallel vector field on a curve a in M. Show that a vectorfield W on a is parallel if and only if it has constant length and the anglebetween V and W is constant.

3. Let a be a curve in M Ã R3. If Y is a vector field on a tangent to M,prove this analogue of Lemma 3.8:

where Y· denotes the Euclidean derivative of Y. Hence the Euclidean accel-eration of a is a = a ≤ + (S(a ¢) •a ¢)U.

4. Let x be the geographical parametrization of a sphere S of radius r, andlet a be the circle of latitude

(In view of the symmetry of S this is a typical circle on S .)(a) Show that a has intrinsic acceleration

where

(b) Compute separately the Euclidean acceleration of a and S(a ¢). Thencheck the last formula in Exercise 3.

5. (Curvature and Holonomy.) Let a be a closed curve in a geometric surface M.(a) If a is the boundary curve of a smooth oriented disk D in M, show

that the holonomy angle ya of a is (Hint: There is always a frame

field on D.)

KdM.DÚÚ

E Gv2 = x / .

¢¢ =a r v v Ecos sin0 0 2,

a pu u v u( ) = ( )x , , for0 0 2� � .

˙ •Y Y S Y U= ¢ + ¢( )( )a ,

b t ct st t st( ) = ( ) >, for all such that ,0

w q12 1= =du v/ .


(b) With notation as in (a), deduce the following characterization of theGaussian curvature of M at a point p:

6. If there is a nonvanishing vector field W on M such that —VW = 0 for allV, show that M is flat. (Hint: There is a frame with .)

7. (Isometries preserve covariant derivatives.) For an isometry F: M Æ N,prove the following two cases:

(a) If V and W are vector fields on M and and are their transferredvector fields on , then

(Hint: By the linearity of — it suffices to assume W = fE1 for a frame fieldE1, E2. Strictly speaking, the transferred vector field is (F* (V))(F -1), butto avoid clutter, write simply = F* (V).)(b) If Y is a vector field on a curve a in M, then

where is the vector field F* (Y) on the curve = F(a) in . Hence inparticular, acceleration is preserved: F* (a ≤) = F(a)≤.This is the analogue of the Euclidean result, Corollary 4.1 of Chapter 3.

8. Let x be an orthogonal patch in a geometric surface M. Prove thatxuv = xvu by first showing that for the associated frame field ,

we have:

(a) . (Hint: Use the (*) property of —.)

(b) . (Hint: Use the formula for w12 in Sec. 6 of Ch. 6.)

9. (Continuation.) To show that yuv = yvu for an arbitrary patch, write y(u, v) = x( (u, v), (u, v)) with x orthogonal, and compute yuv - yvu.

7.4 Geodesics

Geodesics in an arbitrary geometric surface generalize straight lines inEuclidean geometry. We have seen that a Euclidean straight line g (t) = p + tqis characterized infinitesimally by vanishing of acceleration; thus

vu

w12xx x

vvu v

EG=

· , Ò

w12xx x

vuv v

EG=

· , ÒE Gv2 = x /

E Eu1 = x /

MaY

F Y Y* ,¢( ) = ¢

VV

— = —V VW W .

MWV

E W c1 = /

Kp

p( ) =Æ

lim .D D

y a

area


4.1 Definition A curve g in a geometric surface is a geodesic provided itsacceleration is zero, g ≤ = 0.

In other words, the velocity g ¢ of a geodesic is parallel: Geodesics neverturn. In particular, since g parallel implies ||g ¢|| constant, geodesics have con-stant speed.

Acceleration is preserved by isometries (Exercise 7 of Section 3), so itfollows that geodesics are isometric invariants. In fact, if F: M Æ N is merelya local isometry, then F carries each geodesic g of M to a geodesic F(g) of N.(This follows since F is an isometry on small enough neighborhoods.)

For a surface M in R3, a curve in M was defined to be a geodesic if itsEuclidean acceleration is always normal to M. This is consistent with thegeneral definition above, for by Exercise 3 of Section 3 the intrinsic acceler-ation of a curve in M Ã R3 is the component tangent to M of its Euclideanacceleration. Thus the former is zero if and only if the latter is normal to M.

Now we find coordinate conditions for a curve a in a geometric surface Mto be a geodesic. If E1, E2 is a frame field on M, then throughout this sectionwe write the velocity and acceleration of a as

This means that a is a geodesic if and only if A1 = A2 = 0. (Note that v1, v2,A1, and A2 are real-valued functions on the domain I of a.)

From Section 3 these components of acceleration can be expressed in termsof the connection form of E1, E2 as

Now we describe A1 and A2 in coordinate terms.

4.2 Theorem Let x be an orthogonal coordinate patch in a geometricsurface M. A curve a(t) = x(a1(t), a2(t)) is a geodesic of M if and only if

Note the symmetry of these two equations under the reversals 1 ´ 2,u ´ v, E ´ G. In this context we always understand that the functions E and

A aG

E a G a a G av u v2 2 12

1 2 221

22 0= ¢¢ - ¢ + ¢ ¢ + ¢( ) =+ .

A aE

E a E a a G au v u1 1 12

1 2 221

22 0= ¢¢ ¢ + ¢ ¢ - ¢( ) =+ ,

A v v

A v v

1 1 2 21

2 2 1 12

= ¢ + ¢( )= ¢ + ¢( )

w a

w a

,

.

¢ = + ¢¢ = +a av E v E A E A E1 1 2 2 1 1 2 2and .

7.4 Geodesics 347

G and their partial derivatives Eu, Ev, . . . are evaluated on (a1, a2) and hencebecome functions on the domain I of a.

Proof. The velocity of a is a ¢ = a ¢1xu + a ¢2xv, so in terms of the associ-ated frame field of x (Chapter 6, Section 6),

Thus the acceleration components A1, A2 become

(1)

The formula for w12 in Section 6 of the Chapter 6 gives

(2)

When this is substituted into (1), we get

(3)

Standard calculus computations transform (3) into the form stated in theproposition. We remind the reader that in a Leibnizian expansion such as

the notation E is short for E(a1, a2), so

◆

4.3 Theorem Given a tangent vector v to M at a point p, there is a uniquegeodesic g, defined on an interval I around 0, such that

g g0 0( ) = ¢( ) =p v, and .

¢ = ¢ + ¢E E a E au v1 2.

Ea a E aE

E¢( )¢ = ¢¢ + ¢

¢1 1 1

2 ,

A Ea E a aG

Ea

G

A Ga G a aE

Ga

E

vu

uv

1 1 1 2 22

2 2 1 2 12

= ¢( )¢ + ( ) ¢ ¢ -( )

¢

= ¢( )¢ + ( ) ¢ ¢ -( )

¢

.

,

w a w12 12 1 2 1 2¢( ) = ¢ + ¢( ) = -( )

¢ +( )

¢a aE

Ga

EaG

u vv ux x

.

A Ea a G

A Ga a E

1 1 2 21

2 2 1 12

= ¢( )¢ + ¢( ) ¢( )

= ¢( )¢ + ¢( ) ¢( )

.

w a

w a

,

¢ = ¢( ) + ¢( )a Ea E a EG1 1 2 2 .


Thus there are plenty of geodesics in any geometric surface—each deter-mined by its initial position and velocity. In R2, for example, the geodesicdetermined by v at p is g (t) = p + tv.

Proof. Let x be an orthogonal patch around p = x(u0, v0), and write v = cxu + dxv. The geodesic equations in Theorem 4.2 have the form

(1)

for differentiable functions f2 and f2. Furthermore, the initial conditions inthe present theorem are equivalent to

(2)

The fundamental existence and uniqueness theorem of differential equa-tions theory asserts that there is an interval I around 0 on which uniquefunctions a1, a2 are defined that satisfy (1) and (2). Thus g = x(a1, a2) is theunique geodesic defined on I such that g (0) = p and g ¢ (0) = v. ◆

This result is not entirely satisfactory since the interval I may be unneces-sarily small. However, it is easy to make it as large as possible. Suppose g1: I1 Æ M and g2: I2 Æ M are geodesics that both satisfy the initial condi-tions in the theorem. The uniqueness theorem for differential equationsimplies that g1 = g2 on the intersection of I1 and I2. Applying this consistencyresult to all such geodesics gives a single maximal geodesic g : I Æ M satis-fying the initial conditions. (The interval I is the largest possible.) Intuitively,this simply means we let the geodesic run as far as it can.

We ordinarily assume that geodesics are maximal, and denote by gv the onewith initial velocity v.

4.4 Definition A geometric surface M is complete provided every maxi-mal geodesic in M is defined on the whole real line R.

Briefly: geodesics run forever. A constant curve is trivially a geodesic, butexcluding this case, every geodesic has constant nonzero speed. Thus com-pleteness is equivalent to the requirement that all nonconstant geodesics haveinfinite length—in both directions. For example, R2 is certainly complete, andthe explicit computations in Example 6.9 of Chapter 5 show that spheres inR3 are complete as well. More generally, all compact geometric surfaces are

a u a c

a v a d

1 0 1

2 0 2

0 0

0 0

( ) = ¢( ) =

( ) = ¢( ) =

, ,

, .

¢¢ = ¢ ¢( )¢¢ = ¢ ¢( )

a f a a a a

a f a a a a

1 1 1 2 1 2

2 2 1 2 1 2

, , , ,

, , ,

7.4 Geodesics 349

complete (Ch. 8), as are all surfaces in R3 of the form M: f = c. Removal ofeven a single point from a complete surface destroys the property, since geo-desics that formerly passed through the point will be obliged to stop.

It is rarely possible to find explicit solutions to the geodesic differentialequations A1 = 0, A2 = 0 in Theorem 4.2, but the situation can often beimproved by the following result, which shows that either one of these equations can be replaced by a simpler equation. For definiteness, we willreplace A2 = 0.

4.5 Lemma Let E1, E2 be a frame field, and let a be a constant speedcurve such that a ¢ and E2 are never orthogonal. If A1 = 0, then A2 = 0; hencea is a geodesic.

Proof. Taking the derivative of ·a ¢, a ¢Ò = const gives ·a≤, a ¢Ò = 0. Hence

By hypothesis, A1 = 0 and ·E2, a ¢Ò π 0, so A2 = 0. ◆

In coordinate terms this means that the rather complicated second-orderdifferential equation A2 = 0 in Theorem 4.2 has been replaced by the first-order differential equation.

Note that by continuity the lemma remains valid if a ¢ and E1 are orthogo-nal at isolated points.

There is a Frenet theory of curves in a geometric surface that generalizesthat for curves in the Euclidean plane (Exercise 8 in Section 2.3). Since Mhas only two dimensions, torsion cannot be defined. But when M is oriented,curvature can be given a geometrically significant sign as follows.

If b: I Æ M is a unit-speed curve in an oriented surface, then as usual,T = b ¢ is the unit tangent vector field of b. But to get the principal normalvector field of b, we “rotate T through +90°,” defining N = J(T), where J isthe rotation operator in Exercise 3 of Section 1 (see Fig. 7.7). Then the geo-desic curvature kg of b is the unique real-valued function on I for which theFrenet formula

holds. Explicitly, kg = ·T ¢, N Ò. Thus kg is not restricted to nonnegative valuesas in the case of curves in R3: kg > 0 means that T—hence b—is turning in

¢ =T Ngk

Ea Ga¢ + ¢ =12

22 const.

0 1 1 2 2 1 1 2 2= + ¢ = ¢ + ¢A E A E A E A E, , ,a a a .


the positive direction (that is, toward N = J(T)), while kg < 0 means negativeturning (toward -N).

4.6 Corollary Let b be a unit-speed curve in a region oriented by a framefield E1, E2. If j is an angle function from E1 to b ¢ along b, then

Proof. Setting Y = T and a = b in Lemma 3.4 gives

Since J(T) = N, the result follows by comparison with T ¢ = kgN. ◆

In R2 the natural frame field has w12 = 0, so j is the usual slope angle ofthe curve b. Then the result above—called Liouville’s formula—reduces to

In R2 this is often taken as the definition of curvature (see Exercise 8 of Section 2.3).

For an arbitrary-speed regular curve a in M, the present modest Frenetapparatus is defined—just as in Section 4 of Chapter 2—by reparametriza-tion. The same proof as before shows

(*)

where v = ||a ¢|| is the speed function of a.

4.7 Lemma A regular curve a in M is a geodesic if and only if a has constant speed and geodesic curvature kg = 0.

Proof. In (*), since v > 0, we have a ≤ = 0 if and only if ◆dvdt g= =k 0.

¢ = ¢¢ = +a a kvTdvdt

T v Ng, ,2

k jg d ds= / .

¢ = ¢ + ¢( )( ) ( )T J Tj w b12 .

kj

w bg

dds

= + ¢( )12 .

7.4 Geodesics 351

FIG. 7.7

The equations (*) also show that a regular curve a has geodesic curvaturezero if and only if its velocity a ¢ and acceleration a≤ are always collinear. Suchcurves a are sometimes called geodesics; to get a geodesic in the strict senseof Definition 4.1, it suffices to reparametrize a to give it constant speed. Incontexts where parametrization may be important, we call a curve with kg = 0 a pregeodesic.

4.8 Remark Abbreviations. In computations involving curves, wherethere are fixed coordinates, say u and v, it is often convenient to use abbrevi-ations such as u(t) for a1(t) = u(a(t)), and hence u¢(t) for a¢1(t). This is helpfulin computing with differential equations—in particular, with the geodesic differential equations.

Exercises

1. Show that a reparametrization t Æ a( f(t)) of a nonconstant geodesic ais again a geodesic if and only if f has the form f(t) = at + b.

2. If gv is the unique geodesic in M with initial velocity v, show that for anynumber c, gcv (t) = gv (ct) for all t.

3. Find the routes of the geodesics in the stereographic sphere in (1) ofExample 2.4. (Hint: No computation is needed.)

4. Let p1 and p2 be points on the equator of a sphere S. Let b be the closedcurve that starts at the north pole n, follows a meridian to p1, the equator top2, then a meridian back to n. Prove that the holonomy angle of b is the angleat n between the two meridians.

5. (Closed geodesics.)(a) If the boundary curve b: [a, b] Æ M of a smooth disk D is a geodesic,

prove is an integer multiple of 2p. (Sec. 7 shows that it is

exactly 2p.)(b) Deduce that there are no smoothly closed geodesics (equivalently, peri-odic geodesics) on the following surfaces. Use this version of the Jordancurve theorem: Any piecewise-smooth closed curve in a simply connectedsurface is the boundary of a simple region.

(i) A simply connected surface with K � 0.(ii) A paraboloid of revolution.

6. In the projective plane P(r) of radius r (Example 2.7), prove:(a) The geodesics are simple closed curves of length pr.

KdMDÚÚ


(b) There is a unique geodesic route through any two distinct points.(c) Two distinct geodesic routes meet in exactly one point.(Hint: The projection F: S(r) Æ P(r) is a local isometry.)

7. At each point p of a geometric surface M there is an e > 0 such that everygeodesic starting at p runs for at least length e. This follows from differentialequations theory (see Sec. 1 of Ch. 8).

(a) Deduce that in a connected surface M, any two points can be joinedby a broken geodesic segment (that is, a piecewise differentiable curvewhose smooth subsegments are geodesic). (Hint: Use the open set criterionfor connectedness, Ex. 9 of Sec. 4.7.)(b) Give an example of a connected surface M for which broken geodes-ics with at least three geodesic subsegments will be needed in order toconnect all pairs of points. Find M such that arbitrarily many breaks willbe necessary to connect all pairs.

8. Let a be a curve with speed function v > 0 in an oriented surface M.

(a) Show that the geodesic curvature kg of a is

(b) Deduce that if M is an oriented surface in R3, then

where J = –(B, U). Here k is the curvature and B the binormal vector ofa as a curve in R3.

7.5 Clairaut Parametrizations

We consider a special situation where extensive information about geodesicscan be obtained with a minimum of computation.

5.1 Definition A Clairaut parametrization x: D Æ M is an orthogonalparametrization for which both E and G depend only on u, that is, F = 0 and Ev = Gv = 0.

For example, the usual parametrization of a surface of revolution is ofthis type.

5.2 Lemma If x is a Clairaut parametrization, then

(1) All the u-parameter curves of x are pregeodesics, and(2) A v-parameter curve u = u0 is a geodesic if and only if Gu(u0) = 0.

ka a

kg

Uv

=¢ ¥ ¢¢

=•

cos ,3

J

¢¢ ¢( )a a, Jv3

7.5 Clairaut Parametrizations 353

Proof. For (1) it suffices by a remark in Section 4 to show that xu and xuu are collinear. Since xu and xv are orthogonal, this is equivalent to ·xv, xuuÒ = 0. The following equations imply this result.

Similarly, for (2) the v-parameter curve u = u0 is pregeodesic if and only if

The following equations show that this is true if and only if Gu(u0) = 0:

(Recall that xuv = xvu. ) So far, we have not used the condition Gv = 0; itseffect is to show that the v-parameter pregeodesics are in fact geodesics,since it means that they have constant speed. ◆

In the case of a surface of revolution, for example, this lemma providesanother proof that the meridians are geodesics and that a parallel u = u0 isgeodesic if and only if h¢(u0) = 0. (See Exercise 3 of Section 5.6.)

A surface M that has a Clairaut parametrization need not be a surface ofrevolution, but we continue to call its u-parameter geodesics meridians. Ignor-ing the parametrization of these meridians, we can think of them as fibers ofwhich M is composed—and measure the behavior of arbitrary geodesics relative to them.

5.3 Theorem Let a = x(a1, a2) be a unit-speed geodesic with x a Clairautparametrization. If j is the angle from xu to a ¢, then the function

is constant along a. Hence a cannot leave the region where G � c2.

We call the constant c = c(a) thus associated with each geodesic a the slantof a since—in combination with G—it determines the angle j at which acuts across the meridians of x (see Fig. 7.8).

Proof. Because Ev = Gv = 0 for a Clairaut parametrization, the geodesicequation A2 = 0 of Theorem 4.2 reduces to

c G a a G a= ( ) ¢ = ( )1 2 1 sin j

0

20 0 0 0

= = +

( ) = ( ) = ( ) ( )F x

G u G u v u v u v v

v uv v u vv

u u vu v

x x x

x x

, ,

, , , for all

,

, .

x xvv uu v u v0 0 0, , ,( ) ( ) = .

0 2

0

= = =

= = = +

E

F

v u u v u uv

u u v u uu v u vu

x x x x

x x x x x x

, , ,

, , , .


This is equivalent to the constancy of c = Ga¢2, since

To show that sinj, compare the two equations

It follows immediately from |sinj| � 1 that G � c2. ◆

A geodesic with c = 0 has j = 0 or p, and hence it is one of the geodesicmeridians, so we may as well assume c π 0. Moving along a in a direction inwhich G is increasing, the meridians are spreading apart and the constancyof c2 = G(a1) sin2 j shows that |sinj | is decreasing, thus forcing a to turn moretoward the direction of the meridians. On the other hand, if G is decreasingalong a, then a cuts across the meridians at ever-increasing angles. Note thatsin j cannot change sign, for if it is zero even at a single point, then the geo-desic a is tangent to the meridian; hence by the uniqueness of geodesics itparametrizes that meridian.

Such considerations give remarkable information about the general globalbehavior of geodesics in a Clairaut parametrization.

¢ = ¢ + ¢ =

¢ = ¢ -ÊËÁ

ˆ¯ =

a

a ap

j j

, , ,

,

x x x x

x x

v u v v

v v

a a c

G

1 2

2cos sin .

c G=

¢ = ¢( )¢ = ¢ ¢ + ¢¢ = ¢ ¢ + ¢¢c Ga G a Ga G a a Gau2 2 2 1 2 2.

¢¢+ ¢ ¢ =aGG

a au2 1 2 0.


FIG. 7.8

5.4 Example (Global trajectories.) What happens to the geodesicspassing through an arbitrary point p0 on the paraboloid of revolution M:z = x2 + y2? The parametrization x(u, v) = (ucosv, u sinv, u2)—for which u isdistance to the z axis—is Clairaut with E = 1 + 4u2, F = 0, G = u2. Thus theslant of a unit-speed geodesic a (t) = x(a1(t), a2(t)) is

where j(t) is the oriented angle from xu to a ¢(t) (see Fig. 7.9).Suppose a starts upward from p0, that is, with initial angle 0 < j0 < p/2 (by

symmetry, we may as well take j � 0). Then a1 increases; hence |sin j |decreases, so a ¢ turns steadily toward xu, and a escapes to z = •.

If j0 = p, then a parametrizes a meridian of this surface of revolution, soa runs directly down through the vertex 0 of M and back up to z = •.

The interesting case is p/2 < j0 < p. Now a starts downward, but it neverreaches 0—only meridians can do that. It is easy to predict at the start howlow a gets. Initially, since u = a1 is decreasing, |sinj | is increasing, so a ¢ isturning away from the meridians. These changes continue until j reaches p/2.This will occur precisely when c = u0 sinj0 = u1, hence at height

The geodesic a cannot asymptotically approach the nongeodesic v-parameter curve at this height, any more than a straight line in the plane can asymptotically approach a circle. So a meets and bounces off the v-parameter curve (Fig. 7.9) and turning steadily back toward xu, escapes to z = •, as does every geodesic on M.

Not only does the notion of slant give remarkable qualitative informationabout where geodesics can go in a Clairaut surface, but more than that, it

z u umin sin .= = >12

02

02 0j

c G a t t a t t= ( )( ) ( ) = ( ) ( )1 1sin sinj j ,


FIG. 7.9

leads to a single integral formula for pregeodesics. Thus the routes followedby the geodesics are specified—though not their geodesic parametrizations.

5.5 Proposition If x is a Clairaut parametrization, then every geodesica such that a ¢ is never orthogonal to meridians can be parametrized as b(u) = x(u, v(u)), where

with c the slant of a. Hence, by the fundamental theorem of calculus,

Proof. Since a has unit speed,

By Theorem 5.3,

(C-2)

Now substitute (C-2) into the preceding equation, and solve for

(C-1)

The nonorthogonality condition means that a¢1 is never zero on thedomain I of a. Thus, as in elementary calculus, the functions u = a1(s),v = a2(s) can be reparametrized by s = (a1

-1)(u) to give u = u, v = v(u). Then(C-1) and (C-2) give

as required. ◆

In the Clairaut case, one can check that equations (C-1) and (C-2) aboveare not only necessary but also sufficient for a = x(a1(s), a2(s)) to be a unit-speed geodesic. Note that dropping the plus-or-minus sign from (C-1) losesonly a reverse parametrization, since c can have either sign. These equations

dvdu

a sa s

c E

G G c=

¢( )¢( ) = ±

-2

12

.

¢ = ±-

aG c

EG1

2

.

¢ = ( )ac

G a21

.

1 1 12

1 22= ( ) ¢ + ( ) ¢E a a G a a .

v u v uc E dt

G G cu

u

( ) = ( ) ±-Ú0

20

dvdu

c E

G G c= ±

- 2,


are convenient for numerical computation (see Exercises 9–12), although theyfail at turning points since there G = c2. They are, of course, much simplerthan the second-order differential equations in Theorem 4.2.

5.6 Example Routes of geodesics.(1) The Euclidean plane. To illustrate the preceding proposition, we

find the routes of the well-known geodesics of R2 in terms of a polar parametrization

Since E = 1, F = 0, G = u2, this is a Clairaut parametrization. The u-parameter geodesics are just straight lines through the origin. By Proposition5.5, all the others can be parametrized as b(u) = x(u, v(u)), where

Hence or equivalently, ucos(v - v0) = c, which is thepolar equation of a straight line. Here the slant has geometric significance asthe shortest distance from the origin to the line, and the point on the linenearest the origin is a turning point.

(2) The hyperbolic plane. Again we try the polar parametrization. The ruler function h that describes the conformal geometric structure ofH depends only on Euclidean distance u to the origin 0; in fact,

Thus

As in (1), x is a Clauraut parametrization, so the u-parameter curves—Euclidean lines through the origin—are the routes of geodesics. By Pro-position 5.5, the routes of all other geodesics are parametrized by b(u) = x(u, v(u)), where

(1)

This equation can be integrated explicitly by setting

wau

ua

c

c= +Ê

Ëˆ¯ =

+1

4 1

2

2, where .

dvdu

ch u

ch u=

±- ( )

/

/.

2

1

Eh

F Guhu u v v= = = = =x x x x, , , ,

10

2

2

2.

h u h u v u( ) = ( )( ) = +x , / .1 42

v v c u- = ± ( )-0

1cos / ,

dvdu

c

u u c

ddu

cu

=±

-= ± Ê

Ëˆ¯

-

2 2

1cos .

x u v u v u v, ,( ) = ( )cos sin .


Then a straightforward computation gives

(2)

Hence v - v0 = ±cos-1 w, that is,

which we rearrange as

(3)

The Euclidean law of cosines, A2 + B2 - 2ABcosv = C 2, applied to a diagramlike Fig. 7.10, shows that the polar equation of a circle of radius r with centerat x(u0, v0) is

Comparison with equation (3) shows that the route of b is a Euclidean circleC with u0 = r 2 + 4. Since u0 > 2, the center of the circle lies outside the hyper-bolic plane H: x2 + y2 < 4. As can be seen from Fig. 7.10, the relation u0 = r2 + 4 implies that C is orthogonal to the rim x2 + y2 = 4 of H. Of course,b is restricted to the open arc of C inside the rim.

Conclusion: The routes of the geodesics of the hyperbolic plane are the por-tions in H of (1) all Euclidean straight lines through the origin 0 and (2) allEuclidean circles orthogonal to the rim of H (Fig. 7.11). ◆

u u u u v v r202

0 022+ - -( ) =cos .

uu

av v2

044

0+ - -( ) =cos .

cos v v wau

u-( ) = = +Ê

Ëˆ¯0

2

14

,

dvdu

dw du=

±-

/.

1 2w


FIG. 7.10

As these examples show, the geodesics of the hyperbolic plane bear com-parison with those of the Euclidean plane. Around 300 .., Euclid estab-lished a remarkable set of axioms for the straight lines of his plane. The goalwas to derive its geometry from axioms so overwhelmingly reasonable as to be “self-evident.” The most famous of these is equivalent to the parallelpostulate: If p is a point not on a line a, then there is a unique line b throughp that does not meet a.

From the beginning this postulate was regarded as somewhat less certainthan the others. For example, the axiom that two points determine a uniquestraight line might be checked by laying down a (perhaps long, but still finite)straightedge touching both points. But for the parallel postulate, one wouldhave to travel the whole infinite length of b to be sure it never touches a.Thus, over the centuries, tremendous efforts were expended in trying todeduce the parallel postulate from the other axioms. The hyperbolic plane Hoffers the most convincing proof that this cannot be done. For if “straightline” is replaced by “route of a geodesic,” then every Euclidean axiom holdsin H except the parallel postulate. For example, given any two points it is easyto see that one and only one geodesic route runs through them. But it is clearfrom Fig. 7.11 that in H there are an infinite number of geodesic routesthrough p that do not meet a.

When the implications of this discovery were worked out, what wasdestroyed was not the modest hope of proving the parallel postulate, but thewhole idea that the Euclidean plane is, in some philosophical sense, anAbsolute, whose properties are self-evident. It had become just one geomet-ric surface among the infinitely many discovered by Riemann.


FIG. 7.11

Exercises

1. In the Poincaré half-plane, show that the routes of geodesics are all vertical lines and all semicircles with centers on the u-axis (see Fig. 7.12).(Hint: x (u, v) = (u, v) is a Clairaut patch “relative to v,” so in the text equa-tions, reverse u and v, and E and G.)

2. (Barrier curves.) Let x be a Clairaut parametrization, and let a = x(a1, a2) be a unit-speed geodesic with slant c. Suppose that a starts atthe point p0 = x(u0, v0) and, for definiteness, that a¢1(0) > 0. If there is a number u > u0 such that G(u) = c2, let u1 be the smallest such number. Then the v-parameter curve b(v) = x(u1, v) is called a barrier curve for a. Prove:

(a) a comes arbitrarily close to b.(b) If b is a geodesic (that is, if Gu(u0) = 0), then a does not meet b—henceasymptotically approaches it. (See Ex. 11.)

3. (Continuation.) If the barrier curve is not a geodesic, it can be shown thata does meet b, say at the point a(t*). Prove that a¢(t*) = 0 and that a¢ changessign at t*. Thus a has a turning point at a(t*), bouncing off b as in Fig. 7.13.(Hint: Show that a≤(t*) π 0.)

In Example 5.6, what are the barrier curves for the geodesics of R2? thegeodesics of H?


FIG. 7.12

FIG. 7.13

4. Let a be a geodesic on a surface of revolution.(a) Show that the slant of a is c = h sinj, where h(t) is the distance to theaxis of revolution, and j gives the angles at which a cuts the meridians.(b) Deduce that a cannot cross a parallel of radius |c|.

5. On a torus of revolution T with the usual parametrization, prove:(a) If a geodesic a is, at some point, tangent to the top circle (u = ),then a remains always on the outer half of T ( � u � ) and travelsaround T, oscillating between the top circle and bottom circle.(b) Every geodesic of T that crosses the inner equator (u = -p) also crossesthe outer equator (u = 0), and furthermore, unless it is a meridian, it willspiral around the torus, crossing both equators infinitely many times.(c) Every geodesic of T—except a parametrization of the inner or outerequator—crosses the outer equator.

6. Let M be the catenoid in Example 7.1 of Chapter 5, with c = 1. Let Cbe its central circle x = 0. If a is a unit-speed geodesic starting at a typicalpoint, say p0 = (u0, cosh u0, 0), u0 > 0, let j(t) be the angle from the positivedirection on meridians to a ¢(t). In each of the following cases, find initialvalues 0 � j0 � p of j (if any) such that a:

(a) stays in the region u > u0.(b) starts toward the central circle C, but turns back before reaching it.(c) asymptotically approaches C.(d) bounces off C and returns to approach x = +•.(e) crosses C at least twice.(f) crosses C once and continues to x = -•.(Hint: Compare with Ex. 4.)

7. Prove that no geodesic on the bugle surface (Example 7.6 of Ch. 5) canbe defined on the whole real line.

It is often convenient to regard a point of R2 as a complex number

Thus the hyperbolic plane can be described as the disk |z| < 2 with conformal structure given by the ruler function Here |z| is themagnitude of z, characterized by

8. (The Poincaré half-plane P is isometric to the hyperbolic plane H.) Interms of complex numbers, P is the half-plane Iz > 0, with conformal struc-ture given by h(z) = Iz. (Iz is the imaginary part v of z = u + iv.) Let F: H Æ P be the mapping

z zz u v2 2 2= = + .

h z z( ) = -1 42 / .

z u iv u v= + = ( ), .

p /2-p /2p /2


Show that:(a)(b) F is a diffeomorphism of H onto P. (Compute F -1 explicitly.)(c) Relative to Euclidean geometry, F is conformal, with scale factor

(See Ex. 7 of Sec. 1.)(d) F: H Æ D is an isometry.

Make a sketch of H and P indicating the images in P of each of the fourquadrants of H. (Hint: the u and v axes in H are the routes of geodesics.)

9. (Numerical integration, computer graphics.) (a) Let x(u, v) be a Clairautparametrization with metric components E(u), G(u). Write the computercommands that produce:

(a) Numerical solution of the first-order geodesic equations (C-1) and (C-2) in the proof of Proposition 5.5 (changing a1 to u(t), etc.). Use initialconditions u(0) = u0, v(0) = v0, with slant c, on the interval tmin � t � tmax.(b) A plot of the geodesic given in (a).(Dropping the ± in (C-1) costs only a reverse parametrization. The differ-ential equations behave badly when the geodesic meets a barrier.)

10. (Continuation.) Let x(u, v) = (ucosv, u sinv) be the Clairaut parame-trization of the hyperbolic plane H(1) as in Example 5.6.

(a) Plot the geodesic that starts at u(0) = 1, v(0) = 0 with slant c = 1.330(avoiding the turning point problem of c = 4/3) and runs to near the rimof H.(b) Show the following on a single plot: the geodesic of (a), its other branchwith c = -1.330, the relevant arc of its barrier curve u = 1, and the rim u = 2. (Hint: See Fig. 7.11.)

11. (Continuation.) Let M be the surface of revolution with parametriza-tion x(u, v) = (u, f(u) cosv, f(u) sinv), where

(a) Plot the zone of M between u = -5 and u = 2.(b) Using Exercise 9, plot the geodesic g that starts at x(-4, 0) and asymp-totically approaches the neck u = 0 of M on an interval 0 � t � b, whereb is 20 or more.

12. (Continuation.) On the paraboloid given by x(u, v) = (ucosv, u sinv, u2),let g be the geodesic that has a turning point (meets its barrier curve) at u(0) = 1, v(0) = 0. Plot:

(a) The xy projection (ucosv, u sinv) of one branch of g, defined on aninterval 0 � t � b, with b large.(b) On a single figure, the branch in (a) and its symmetrical branch.

f u u u( ) = +( ) +( )3 42 2/ .

l z iz( ) = +4 2 2/ .

IF z z iz( ) = -( ) +4 22 2/ .

F zz iiz

z( ) =++

<( )22

2 .


13. A Liouville parametrization x: D Æ M is an orthogonal parametriza-tion for which E = G = U(u) + V(v). (Thus Clairaut parametrization is thespecial case where U or V is zero.) If a = x(a1, a2) is a unit speed geodesic,with x Liouville, prove that

is constant along a, where j is the angle from xu to a ¢.(Hint: First show that sin2 j = (U(a1) + V(a2))a¢12 and cos2 j =

(U(a1) + V(a2))a¢22.)

14. (Continuation of Exercise 8.) (a) Show that the restriction to H of any orthogonal transformation of R2 (e.g., a Euclidean rotation about 0) is anisometry of H, and that for any real number a, the Euclidean translation Ta (z) = z + a is an isometry of P. (Hint: See Ex. 9 of Sec. 1.)

(b) If F: H Æ P is the isometry in Exercise 8, show by explicit computtionthat the isometry F -1TaF: H Æ H carries 0 to a point wa in H with

(c) Deduce that given any point w of H there is an isometry of H thatsends 0 to w.

Thus, all points of the hyperbolic plane are geometrically equivalent. A mildextension of (c) shows that all frames are geometrically equivalent.

7.6 The Gauss-Bonnet Theorem

We have seen that the Gaussian curvature K of a geometric surface M has astrong influence on other properties of M, notably the shape of M when itis a surface in R3. Now we will show that the influence of Gaussian curva-ture penetrates to the topological conformation of M—to properties inde-pendent of the geometry of M.

To show this, the main step is a theorem that relates the total curvature ofa 2-segment to the total amount its boundary curve turns. The geodesic cur-vature of a curve a in an oriented surface M tells its rate of turning relativeto arc length s. So to find the total turning, we integrate with respect to arclength, adjusting suitably when a is merely a regular curve.

6.1 Definition Let a: [a, b] Æ M be a regular curve segment in an oriented geometric surface M. The total geodesic curvature of a is

k ka

g gs a

s b

ds s tdsdt

dtÚ Ú= ( )( )( )

( ).

w a aa = +( )4 42 2/ .

U a V a12

22( ) - ( )sin cosj j


For example, if a makes one counterclockwise trip around a circle C ofradius r in the plane R2, then a has constant geodesic curvature Thus, regardless of the size of r,

The usual orientation of R2 makes kg > 0 for a (left-turning) counterclock-wise trip, but kg < 0 for a (right-turning) clockwise trip. So a clockwise triparound C would have total curvature -2p.

Integrating the formula in Corollary 4.6 gives

6.2 Lemma Let a: [a, b] Æ M be a regular curve segment in a region ofM oriented by a frame field E1, E2. Then

where j is an angle function from E1 to a ¢ along a, and w12 is the connec-tion form of E1, E2.

For integration we have used 2-segments x: R Æ M that are one-to-oneand regular only on the interior R° of R, since 2-dimensional integration canafford to neglect 1-dimensional sets. But now the boundary of x becomesimportant, so we require x to be one-to-one and regular on the entire rec-tangle R. Equivalently, x: R Æ M is the restriction to R of a patch definedon some open set containing R.

Then the edge curves a, b, g, d of x (Definition 6.4 of Chapter 4) are one-to-one regular curve segments, and the boundary of x,

is a single piecewise regular curve enclosing the rectangular region x(R) Ã M.Now we want to define the total geodesic curvature of ∂ x. If (contrary

to fact) ∂ x were a single smoothly closed curve g, this total would be just

But since ∂ x has corners, it is not enough to get the total turning on

the edge curves, namely,

We must also add the angles through which a unit tangent on ∂ x would haveto turn at the four corners of the rectangular region x(R). These angles replace

k k k k k

k k k k

a b g d

a b g d

g g g g g

g g g g

ds ds ds ds ds

ds ds ds ds

∂ - -Ú Ú Ú Ú ÚÚ Ú Ú Ú

= + + +

= + - -

x

.

kg

gdsÚ .

∂ x = + - -a b g d ,

k j j wa a

gds b aÚ Ú= ( ) - ( ) + 12,

k p pa

gdsr

rÚ = =1

2 2 .

k g r= 1/ .

7.6 The Gauss-Bonnet Theorem 365

the total curvature of small curved segments needed to round off the cornersand make ∂ x a single smooth curve.

For R: a � u � b, c � v � d these corners,

are called the vertices of x(R).In general, if a regular curve segment a ends at the starting point of

another segment b, say a(1) = b(0), then the turning angle e from a to b isdefined to be the oriented angle from a ¢(1) to b ¢(0) that is smallest in absolutevalue. For a 2-segment—if no other orientation has been specified—we usethe orientation determined by x, represented by the area form dM such thatdM(xu, xv) > 0. The following terminology is familiar in the case of a polygonin the Euclidean plane.

6.3 Definition Let x: R Æ M be a one-to-one regular 2-segment with ver-tices p1, p2, p3, p4. The exterior angle ej of x at pj (1 � j � 4) is the turningangle at pj derived from the edge curves a, b, -g, -d, a, . . . in order of occur-rence in x (Fig. 7.14). The interior angle ij at pj is p - ej.

This general definition will be needed later, but in the case at hand, exte-rior angles can be expressed directly in terms of the usual coordinate angle0 < J < p from xu to xv as

where Jj is the coordinate angle at pj. For example, in Fig. 7.15 consider thesituation at p3. By the definition of edge curves, b ¢ is xv, but (-g)¢ is -xu, since-g is an orientation-reversing reparametrization of g. Thus e3 + J3 = p.

e p J e J e p J e J1 1 2 2 3 3 4 4= - = = - =, , , ,

p x p x p x p x1 2 3 4= ( ) = ( ) = ( ) = ( )a c b c b d a d, , , , , , , ,


FIG. 7.14

We can now prove one of the fundamental theorems of differential geometry.

6.4 Theorem Let x: R Æ M be a one-to-one regular 2-segment in a geo-metric surface M. If dM is the area form determined by x, then

where ej is the exterior angle at the vertex pj of x (1 � j � 4).

We emphasize that the 2-segment x supplies the necessary orientation;M itself need not be oriented—or even orientable.

This result is called the Gauss-Bonnet formula with exterior angles. Since ej = p - ij for 1 � j � 4, the formula can be rewritten in terms of the interiorangles of x(R) as

Proof. The associated frame field E2 = J(E1) of x hasdM(E1, E2) = +1. Then the second structural equation becomes

The power for this proof is supplied by Stokes’ theorem (6.5 of Ch. 4),which gives

(1)K dMx xÚÚ Ú+ =w12 0

∂.

d K K dMw q q12 1 2= - Ÿ = - .

E Eu1 = x / ,

KdM dsgX XÚÚ Ú+ = + + + -k i i i i p

∂1 2 3 4 2 .

KdM dsgX XÚÚ Ú+ + + + + =k e e e e p

∂1 2 3 4 2 ,


FIG. 7.15

Now we use Lemma 6.2 to evaluate

(2)

Beginning with a, we have E1, so the angle from E1 to a ¢ isidentically zero. Thus by Lemma 6.2,

(3)

Now we try a harder case, say Here the angle from to

d ¢ = xv is just the coordinate angle J from xu to xv. (See Fig. 7.16.) Henceby Lemma 6.2,

where Jj is the coordinate angle at the vertex pj (1 � j � 4). But since J1 =p - e1 and J4 = e4, this becomes

(4)

In an entirely similar way we find

(5)

and

(6)

Equations (3)–(6) turn (2) into

Substitution in (1) then yields the required formula. ◆

w k k k k

p e e e e

k p e e e e

a b g d12

1 2 3 4

1 2 3 4

2

2

∂

∂

x

x

Ú Ú Ú Ú Ú

Ú

= + - -

- + + + +( )

= - + + + +( )

g g g g

g

ds ds ds ds

ds .

wg g

k12Ú Ú= gds.

wb b

p e e k12 2 3Ú = - + + + Ú gds.

w p e e kd d

12 1 4Ú Ú= - - + gds.

w J J kd d

12 1 4Ú Ú= - + gds,

E Eu1 = x /wd

12Ú .

w ka a

12Ú Ú= gds.

¢ = =a xu E

w w w w wa b g d

12 12 12 12 12∂ xÚ Ú Ú Ú Ú= + - - .


FIG. 7.16

Our goal now is to extend the reach of the Gauss-Bonnet formula to anentire surface.

A rectangular decomposition D of a surface M is a finite collection of one-to-one regular 2-segments x1, . . . , xf whose images cover M in such a waythat if any two intersect, they do so in either a single common vertex or asingle common edge.

Evidently, a rectangular decomposition is a special kind of paving (Defin-ition 7.3 of Chapter 6), but the regions xi(Ri) now have well-defined regularedges. Furthermore, they are required to fit together very neatly, as in Fig.7.17. (Compare the more casual paving in Fig. 6.14.)

6.5 Theorem Every compact surface M has a rectangular decomposition.

This result is certainly plausible, for if M were made of paper, we couldjust take a pair of scissors and cut out rectangular pieces until all of M wasgone. In fact, any compact region bounded by a finite number of piecewiseregular curve segments has a rectangular decomposition. A rigorous proofwould be tedious and not very instructive.

The topological analogue of a diffeomorphism is a homeomorphism, a con-tinuous mapping F: M Æ N that has a continuous inverse map F -1: N Æ M.When such an F exists, M and N are said to be homeomorphic. A topologicalinvariant is a property that is preserved by homeomorphisms, hence sharedby homeomorphic surfaces. These are the properties that can be defined solelyin terms of open sets.

Every diffeomorphism is a homeomorphism, because differentiable func-tions are continuous. But the converse is false; indeed, it is already clear inelementary calculus that a continuous function f: R Æ R can have a jagged


FIG. 7.17

graph, while differentiability requires smoothness. Nevertheless, in dimension2 we have the remarkable result that two surfaces are diffeomorphic if and onlyif they are homeomorphic. This is a considerable simplification, since it allowsmany topological invariants to be discussed in differentiable terms. Here is afamous instance.

We shall understand that a rectangular decomposition D carries with it notonly its rectangular regions xi(Ri)—called faces—but also the vertices andedges of these regions.

6.6 Theorem If D is a rectangular decomposition of a compact surfaceM, let v, e, and f be the numbers of vertices, edges, and faces in D. Then theinteger v - e + f is the same for every rectangular decomposition of M. Thisinteger c(M) is called the Euler characteristic of M.

An elementary topological proof of this famous theorem is outlined inChapter 1, Section 8 of [Ma].

The fact that the decomposition D is based on rectangles is merely a con-venience for integration. We could just as well cut M into arbitrary polygons(see Exercise 5 of Section 6). In the resulting polygonal decomposition, the dif-ferent polygons of course are still required to fit neatly, but (as in Fig. 7.18)they need not have the same number of sides. When only triangles are used,the decomposition is called a triangulation of M.

6.7 Example Euler characteristic.(1) Any sphere S has c(S ) = 2. The surface of a tetrahedron provides a tri-

angulation of S , and we count v = 4, e = 6, f = 4; hence c(S ) = 2. By inflat-ing a cube, as in Fig. 7.18, we get a rectangular decomposition of S with v = 8, e = 12, f = 6, so again c(S ) = 2. Inflating a prism gives a polygonaldecomposition with v = 6, e = 9, f = 5, but still c(S ) = 2.

(2) A torus T has c(T) = 0. Picture T as a torus of revolution, and cut italong any three meridians and three parallels. This gives a rectangular decom-position with v = 9, e = 18, f = 9; hence c = 0.


FIG. 7.18

(3) Adding a handle to a compact surface reduces its Euler characteristicby 2.

This last assertion requires explanation. A handle is a torus with the inte-rior of one face removed. To add a handle to a surface M—also in some rec-tangular decomposition—remove the interior of a face of M, and to theresulting rim, smoothly attach the rim of the handle, with the vertices andedges of the two rims coinciding (Fig. 7.19).

This operation produces a new surface M¢ already furnished with a rec-tangular decomposition. Its Euler characteristic is

In fact, the decomposition has exactly two faces fewer than M and the toruscombined. Coalescing the two rims has also eliminated four vertices and fouredges, but this has no effect on c.

It is easy to see that diffeomorphic surfaces have the same Euler character-istic, for if x1, . . . , xf is a decomposition of M and F: M Æ N, then F(x1),. . . , F(xf) is a decomposition of N with the same v, e, and f.

For example, no matter how wildly we distort the “round” sphere

the resulting surface will still have Euler characteristic 2.Suppose we start from a sphere S and successively add h handles (h � 0)

to obtain a surface S [h]. What the handles look like and where they areattached is irrelevant in view of the following remarkable result.

6.8 Theorem If M is a compact, connected, orientable surface, there isa unique integer h � 0 such that M is diffeomorphic to S [h].

In this case, we say that M itself is a sphere with h handles. The theoremis proved in Chapter 1 of [Ma], where by a remark above, homeomorphismcan everywhere be replaced by diffeomorphism.

In Fig 7.20 the surfaces all have just one handle, so they are all diffeo-morphic. If M has h handles, then by (1) and (3) of the preceding example,c(M) = 2 - 2h, since

S : x y z2 2 2 1+ + = ,

c c¢( ) = ( ) -M M 2.


FIG. 7.19

6.9 Corollary Compact orientable surfaces M and N have the same Eulercharacteristic if and only if they are diffeomorphic.

Proof. If c(M) = c(N), then M and N have the same number of handles;hence by the theorem they are diffeomorphic. The converse was notedabove. ◆

This is the sense in which c(M) is characteristic of M. Though the numberof handles (or genus) of M is visually appealing, c(M) is usually easier tocompute and in the long run turns out to be more fundamental.

Returning to geometric surfaces, we can now prove this spectacular con-sequence of the Gauss-Bonnet formula.

6.10 Theorem (Gauss-Bonnet) The total Gaussian curvature M of acompact orientable geometric surface M is 2p times its Euler characteristic:

Proof. Orient M by an area form dM, and let D be a rectangular decom-position of M whose 2-segments are all positively oriented. (A wrong ori-entation of x can be corrected by reversing u and v.) Then D is an orientedpaving of M as defined in Chapter 6, Section 7. By definition, the total cur-vature of M is

K dM MM

= ( )ÚÚ 2pc .

c c cM h h h( ) = [ ]( ) = ( ) - = -S S 2 2 2 .


FIG. 7.20

(1)

Apply the Gauss-Bonnet formula to each summand. In terms of inte-rior angles, the result is

(2)

Now consider what happens when (2) is substituted into (1).Because M is a surface—locally like R2—each edge of the decomposi-

tion D will occur in exactly two faces, say xi(Ri) and xj(Rj). Let ai and aj

be the parametrizations of this edge occurring in the oriented boundaries∂xi and ∂xj, respectively.

By construction, the regions xi(Ri) and xj(Rj) have the same orientationas M, so ai and aj are orientation-reversing reparametrizations of eachother, as shown in Fig. 7.21(a). Thus

It follows that summing over all faces yields

(3)

In fact, we have just seen that the integrals over edge curves cancel in pairs.

As usual, v, e, and f are the numbers of vertices, edges, and faces in thedecomposition. Substituting (2) into (1) gives

(4)

where I is the sum of all interior angles of all the 2-segments in the decom-position. But the sum of the interior angles at each vertex is just 2p (Fig.7.21(b)), so I = 2pv. Thus

K dM fMÚÚ = - +2p I ,

k gi

f

dsi∂ xÚ

=Â =

1

0.

k ka a

g gds dsi j

Ú Ú+ = 0.

K dM dsi i

gx xÚÚ Ú= - - + + + +k p i i i i

∂2 1 2 3 4 .

K dM K dMM i

f

iÚÚ ÚÚ=

=Â

x1

.


FIG. 7.21

(5)

A simple combinatorial observation will complete the proof. The faces of the decomposition D are rectangular: Each face has four edges. But eachedge belongs to two faces. Thus 4f counts e twice; that is, 4f = 2e. Equiva-lently, -f = f - e, so (5) becomes

◆

The most striking aspect of the Gauss-Bonnet theorem is that it directlylinks topology and geometry. Because the Euler characteristic is a topologi-cal invariant, the theorem shows that total Gaussian curvature is a topologi-cal invariant.

For example, the flat torus in Example 2.2 has K = 0, hence total curva-ture zero. Earlier, an explicit calculation showed that although the (diffeo-morphic) torus of revolution in R3 has variable curvature, its total curvatureis also zero. The Gauss-Bonnet theorem makes this evident without calcula-tion—and if the various tori in Fig. 7.20 were realized formally, they toowould have total curvature zero.

The Gauss-Bonnet theorem provides a way to attack some seemingly for-midable problems. For instance, Example 2.4(1) shows that if a single pointis removed from a sphere S, there exists a metric on the punctured spherewith K = 0. But

There can be no metric on an entire sphere for which K � 0.

Indeed, K � 0 would imply but we know that 2pc(S) > 0.

Reversing this argument,

If a compact orientable geometric surface M has K > 0, then M is diffeo-morphic to a sphere.

Since K > 0, M has positive total curvature, hence positive Euler charac-teristic. But c(M) = 2 - 2h, so h = 0, and hence M is diffeomorphic to S [0] = S .

Further applications of the Gauss-Bonnet theorem are given in the nextsection.

Exercises

1. Find the total Gaussian curvature of:(a) An ellipsoid. (b) The surface in Fig. 4.8.(c) M: x2 + y4 + z6 = 1.

KdSSÚÚ � 0,

K dM v e f MMÚÚ = - +( ) = ( )2 2p pc .

K dM f vMÚÚ = - +2 2p p .


2. Prove that for a compact orientable geometric surface M:

K > 0 fi M is diffeomorphic to a sphere,

K = 0 fi M is diffeomorphic to a torus,

K < 0 fi M is diffeomorphic to a sphere with h � 2 handles.

3. Let M be a compact orientable geometric surface with h handles. Provethat there exists a point p in M at which

4. If M is a compact orientable geometric surface in R3 that is not diffeo-morphic to a sphere, show that there is a point p of M at which K(p) < 0.(Compare Thm. 3.5 of Ch. 6.)

5. In each case, show that the change in polygonal decomposition of acompact surface M does not change the Euler characteristic v - e + f ofM:

(a) Given a rectangular decomposition of M, cut each rectangle along adiagonal to form two triangles, thus producing a triangulation of M.(b) Given a triangulation of M, cut each triangle into three rectangles bylines from a central point to a midpoint of each side, thus producing a rec-tangular decomposition of M.(c) Given an arbitrary polygonal decomposition of a compact surface M,cut each polygon into triangles, thus producing a triangulation of M.

6. (a) For a regular curve segment a: [a, b] Æ M, show that the total

geodesic curvature is

(Hint: Ex. 8 of Sec. 4.)(b) Let x be a positively oriented orthogonal patch in M. Deduce the fol-lowing formulas for the total geodesic curvature of the parameter curves:

Assume for simplicity that M is in R3. Thus

where either intrinsic or Euclidean derivatives give the same result.

x x x xuu v v vv u uE G, , ,= - = -12

12

and

- ( ) ( )Ú Ú12

120 0

1

2

1

2E

EGu v du

G

EGu v dvv

u

uu

u, , ,

n

.

¢¢ ¢( )¢ ¢Ú a a

a aa

,,J

dt.

ka

gdsÚ

K h

K h

K h

p

p

p

( ) > =

( ) = =

( ) <

0 0

0 1

0 2

if ,

if ,

if .�


7. Let x: R Æ S (r) be the restriction of the geographical patch (Ex. 2.2 ofCh. 4) to the rectangle R: 0 � u, v � Check the Gauss-Bonnet formulaby computing separately each of its terms.

8. If F: M Æ N is a mapping of compact oriented surfaces, the degree dF

of F is the algebraic area of F(M) divided by the area of N. Thus dF gives thetotal algebraic number of times F wraps M around N. (It can be shown thatdF is always an integer.)

Prove the Hopf theorem: If M is a compact oriented surface in R3, thedegree of its Gauss map is

9. Consider a polygonal decomposition whose faces are hexagons, withexactly three edges meeting at each vertex. Is there such a decomposition onthe sphere? the torus?

7.7 Applications of Gauss-Bonnet

The Gauss-Bonnet theorem (6.10) was proved by cutting an entire surface Minto rectangular regions and applying the Gauss-Bonnet formula (6.4) toeach. The scheme works because these rectangles are all consistently orientedby an orientation of M, and thus the integrals Úkgds on their boundariescancel in pairs. Here in essence is the fundamental idea of algebraic topol-ogy—specifically, homology theory. (Indeed, considerations of this kind ledPoincaré to its invention.) By applying this scheme to suitable regions in Mwe can extend the range of the Gauss-Bonnet theorem.

7.1 Definition An oriented polygonal region P in a surface M is a (nec-essarily compact) oriented region furnished with a positively oriented rec-tangular decomposition x1, . . . , xf.

A boundary segment of P is a curve segment b that is an edge curve ofexactly one of the rectangles xi(Ri). For simplicity, we add the requirementthat a vertex of the decomposition cannot belong to more than two bound-ary segments.

We will define the boundary of the region P as a generalization of theboundary of a single 2-segment. Each boundary segment s of P is an edgecurve of exactly one rectangle, say x(R), and has an orientation given by thedefinition of the boundary ∂x (Definition 6.4 in Chapter 4). Recall that in

∂ x = + - -a b g d ,

c M( ) / .2

p / .4


the parametrizations that g and d inherit from x determine the “wrong” ori-entation (see Fig. 4.37). The minus signs correct this—as would an orienta-tion-reversing reparametrization.

With the proper orientations, the boundary segments of P comprise a finitenumber of simply closed polygonal curves b1, . . . , bk. As in the case of asingle rectangle, these orientations obey the informal rule, “Travel around theboundary keeping the region always on the left.”

7.2 Definition The oriented boundary ∂P of an oriented polygonal regionP is the formal sum of the simple closed, oriented polygonal curves bi

described above:

This care with orientation is needed to reach the simplicity of the follow-ing generalized Stokes’ theorem.

7.3 Theorem If f is a 1-form on an oriented polygonal region P, then

In particular, if P is an entire compact oriented surface M, then

Proof. By definition,

Then Stokes’ theorem for rectangles (Theorem 6.5 of Chapter 4) gives

(*)

As in the proof of the Gauss-Bonnet theorem, those edges that belong totwo rectangles acquire opposite orientations from them, so the resultingtwo integrals cancel. These double edges can thus be ignored, leavingexactly the edges in the oriented boundary ∂P. Hence

Substituting this into (*) gives the primary result. If P = M, there are noboundary edges, so all edge contributions cancel. ◆

The following result suggests a new meaning for the Euler characteristic.

f f fb∂ ∂xi ji

f

j

k

Ú Ú Ú= =Â Â= =

1 1 P

.

dii

f

f fP

ÚÚ Ú==Â

∂ x1

.

d dii

f

f fP

ÚÚ ÚÚ==

Âx

.1

dM

fÚÚ = 0.

df fP PÚÚ Ú=

∂.

∂ P = + +b b1. . . .k

7.7 Applications of Gauss-Bonnet 377

7.4 Corollary The following properties of a compact orientable surfaceare equivalent.

(1) There is a nonvanishing tangent vector field on M.(2) c(M) = 0.(3) M is diffeomorphic to a torus.

Proof. (1)Æ(2) Let V be a nonvanishing vector field on M. Then for anygeometric structure on M, the associated frame field

is defined on the entire surface. If w12 is its connection form, we know thatdw12 = -KdM. Using both the Gauss-Bonnet theorem and Stokes’ theorem,we find

(2)Æ(3) From the preceding section, c(M) = 0 implies that M has exactlyone handle, and hence is a torus.

(3)Æ(1) Use xu or xv from the usual parametrization of a torus ofrevolution. ◆

The following result generalizes the Gauss-Bonnet formulas and theorem.

7.5 Theorem If P is an oriented polygonal region in a geometric surface,then

where Sej is the sum of the exterior angles (Definition 6.3) of all the closedboundary curves comprising ∂P.

The proof is virtually the same as for the Gauss-Bonnet theorem (6.10),but now the boundary curves survive and give the additional terms. (See Fig. 7.22.)

The simplest case of this theorem occurs when the polygonal region is asingle triangle (hence has Euler characteristic 1).

7.6 Corollary If D is a triangle in an oriented geometric surface M,then

K dM dsgD DÚÚ Ú+ = - + +( ) = + +( ) -

∂k p e e e i i i p2 1 2 3 1 2 3 .

K dM dsg jP P

PÚÚ Ú+ + = ( )Âk e pc∂

2 ,

2 012pc wM K dM dM M

( ) = = - =ÚÚ ÚÚ .

EVV

E J E1 2 1= = ( ),


If the edges of the triangle are geodesics, then of course the geodesic curva-ture terms vanish. If, further, Gaussian curvature K is constant, this resultreduces to

where A is the area of the triangle. Thus the celebrated theorem of planegeometry that the sum of the interior angles of a triangle is p depends on thefact that R2 is flat. Indeed, for constant K π 0, the sum of the angles is deter-mined by the area of the triangle—and vice versa. Fig. 7.23 shows how a geo-desic triangle manages to have i1 + i2 + i3 greater than p on a sphere (K > 0)and less than p on a hyperbolic plane (K < 0).

KA = + + -i i i p1 2 3 ,


FIG. 7.22

FIG. 7.23

Corollary 7.4 linked Euler characteristic zero to the existence of nonvan-ishing vector fields. Now we look for a generalization. On an arbitrarycompact surface M it is always possible to define a vector field that is differ-entiable and nonvanishing except at a finite number of points. If we think ofthe vectors of the field as the velocity vectors of the smooth steady-state flowof a fluid, then these points are singularities, where the flow stops or is tur-bulent. We want to assign to each such point a number, called its index, thatmeasures how bad the singularity is.

A point p is an isolated singular point of a vector field V if V is nonvan-ishing and differentiable on some neighborhood N of p—except at the pointp itself.

For example, on a sphere S Ã R3, let N be the unit tangent vector field thateverywhere points due north. Evidently, there is no way to define V differen-tiably at the poles (0, 0, ±1), so they are isolated singular points. We view thesouth pole (0, 0, -1) as the source of a fluid flowing toward a sink at the northpole (0, 0, 1).

To define the index of an isolated singular point p of V, let D be an ori-ented disk centered at p in a coordinate patch of M and small enough so thatit contains no other singular points of V. Then the behavior of V on theboundary C of D is already enough to tell how bad the singularity is. The ideais to compare V on C with the restriction to C of a smooth vector field Xthat has no singularities anywhere in D (for example, one of the partial veloc-ities of the patch).

Let a: [a, b] Æ C be a parametrization of C as the oriented boundary ∂D

of D. Let j = –a(X, V) be an angle function from Xa to Va (these vector fieldsrestricted to a).

As Va tranverses C, changes in the angle j measure the rotation of Va rel-ative to Xa. After one circuit these vectors return to their initial values; hencethe total rotation j(b) - j(a) is an integer multiple of 2p. For example, inFig. 7.24, where Va and Xa are initially equal, the total rotation is 2p.

7.7 Definition With notation as above, the index of V at p is the integer

We will soon see that this definition depends only on V and p (and the orientation of D), but not on the other choices involved. This can be ob-served experimentally in Fig. 7.25. There, in each case, the general characterof the vector field V near the singularity is described by drawing some ofits integral curves (Exercise 13 of Section 4.8) since V supplies all their veloc-ity vectors.

ind ,Vb a

p( ) =( ) - ( )j j

p2.


Picture X as a unit vector field in the positive x-direction, and count thenumber of counterclockwise rotations of V relative to X on any circle aroundthe central singularity. For example, Fig. 7.25 indicates that sources haveindex +1. Reversing the direction of the vectors in this figure shows that sinkshave the same index, +1.

Indices are just what is needed to generalize Corollary 7.4.

7.8 Theorem (Poincaré-Hopf) Let V be a vector field on a compact ori-ented surface M. If V is differentiable and nonvanishing except at isolatedsingular points p1, . . . , pk, then the Euler characteristic of M is the sum oftheir indices.

c M V ii

k

( ) = ( )=

Â ind , p1

.


FIG. 7.24

FIG. 7.25

Proof. The theorem is topological, but to simplify the proof we canassume, without loss of generality, that M is a geometric surface.

The case k = 0 is just Corollary 7.4, so suppose k > 0. Since the pointspi are isolated, by the Hausdorff axiom there is a coordinate disk Di aroundeach such that no two disks meet. Let M be the polygonal region thatremains when the interiors of all the disks are removed from M. Thus

(1)

Since V has no singularities on M, it has an associated frame fieldE2 = J(E1) on M. If is the connection form of this frame

field, then by Stokes’ theorem,

(2)

The initial minus sign disappears because by the definition of index, eachCi derives its orientation from Di Ã M, so M gives it the reverse orientation.

For each i = 1, . . . , k, let Xi be a differentiable unit vector field definedon the disk Di. Again, for each i, we have a frame field Xi, J(Xi) on the diskDi. Denote their connection forms collectively by w12.

Let gi: [a, b] Æ Ci be a positively oriented parametrization of Ci = ∂Di.By Stokes’ theorem,

(3)

Substituting (2) and (3) into (1) gives

(4)

Now let Pi be a parallel unit vector field along gi. Oriented angles aredetermined only up to addition of a multiple of 2p ; at the initial point awe choose them so that

(5)

By continuity, this relation persists along gi. Write j = –(Xi,Pi) and =–(V, Pi) = - –(Pi, V). Then integrating (5) shows that

(6)

is the total rotation of V relative to Xi along Ci. We saw in Section 3 thatsince Pi is parallel, the integrand here equals (g ¢) - w12(g ¢). The rota-tion term is just 2p ind(V, pi), so (6) becomes

w w pg

12 12 2-( ) = ( )Úi

V iind , p .

w12

¢ - ¢( )Ú j ja

b

dt

j

–( ) + –( ) = –( )X P P V X Vi i i i, , , .

K dMM i

k

iÚÚ Ú= -( )

=Â w w

g12 12

1

.

K dMi iD

ÚÚ Ú= - wg

12.

K dMC

i

k

iM MÚÚ Ú ÚÂ= - =

=

w w12 12

1∂

.

w12E V V1 = / ,

K dM K dM K dMM i

k

iÚÚ ÚÚ ÚÚ= +

=Â

M D1

.


Substituting this into (4) gives

The Gauss-Bonnet theorem then completes the proof. ◆

Thus for any choice of vector field on M with only finitely many singular-ities (necessarily isolated), the sum of their indices is the same—and dependsonly on the topology of M.

To see that the index is independent of the various choices involved in its definition, suppose that in the theorem we use different ingredients D, X, a at one singular point, say p1. In the Poincaré-Hopf equation,

ind(V, pi), every term except ind(V, p1) is unchanged—hence

ind(V, p1) is unchanged.

7.9 Example The Poincaré-Hopf theorem provides an efficient way tocompute Euler characteristics. For instance, we saw earlier that the due northvector field N on the sphere S has two singularities: a source and a sink. Eachhas index +1, so we find again that c(S ) = 2.

Fig. 7.26 shows a top view of a double torus, that is, a sphere with twohandles. A few integral curves are drawn for a vector field V on M with onevisible singularity. There is a symmetrical pattern of curves on the bottomhalf of M, with a corresponding singularity. Both singularities are of themeeting of two streams type, which has index -1. Hence c(M) = -2, as foundin Section 6 by quite different means.

The definition of index makes sense for a point p where the vector field Vis nonsingular, that is, differentiable and nonzero. But then the index is zero,since we can use V itself as the base vector field X in the definition of index;so the angle j is identically zero.

The converse is not true, that is, ind(V, p) = 0 does not imply that V is non-singular at p. But it is almost true, for it can be shown that V can be modi-

x Mi

( ) = Â

K dM VM

ii

k

ÚÚ = ( )=Â2

1

p ind , p .


FIG. 7.26

fied near p so as to eliminate all singularities there and introduce no new oneselsewhere. Thus isolated singular points with index zero are said to be remov-able. From the viewpoint of Theorem 7.8, a vector field whose singularitiesare all removable is as good as one with no singularities at all.

7.10 Remark The vector fields in this section have been assumed to bedifferentiable, but only continuity is actually required. In general, continuousconsequences can often be derived from differentiable theorems since(roughly speaking) continuous functions can always be approximated by dif-ferentiable functions. In Corollary 7.4, for example, we need only assume thatthe nonvanishing vector field V in assertion (1) is continuous, since a suffi-ciently close differentiable approximation will also be nonvanishing.

Exercises

1. If P is an oriented geodesic n-polygon (that is, the sides of P are geo-desics), show that

where ej and ij are the exterior and interior angles of P, respectively.

2. On a surface M with K � 0, prove that there exist no geodesic n-polygons with n = 0, 1, or 2. (In other words, in M, a smoothly closed geo-desic or a broken geodesic with either one or two corners cannot bound asimple region.)

3. On the standard sphere S Ã R3, since geodesics follow great circles theyare simply closed, and any two must meet.

(a) If M is a compact surface with K > 0, prove that any two simply closedgeodesics must meet.(b) Describe an example where M is diffeomorphic to S and K � 0 but theconclusion of (a) fails.

4. (a) If P is a geodesic n-polygon in the Euclidean plane, show that n � 3 and the sum of the exterior angles is 2p, independent of n.

(b) On a sphere S , for which values of n � 0, do there exist closed brokengeodesics with n corners?

5. As in the definition of index, let a parametrize a curve surrounding acommon isolated singularity p of vector fields V and W. Prove that V and Whave the same index at p if either:

K dM nP

jj

n

jj

n

ÚÚ = - = -( ) += =

Â Â2 21 1

p e p i ,


(a) There is a continuous family of vector fields Vs, for 0 � s � 1, nonvan-ishing on a, such that V0 = V and V1 = W, or(b) 0 < –(V, W) < 2p. (In particular, ±V and ±J(V) all have the same index.)

6. In each case, sketch and describe a vector field on the sphere S Ã R3 thathas (a) exactly two singular points, but no sources or sinks; (b) only one sin-gular point (Hint: stereographic projection); (c) six singular points (index zeronot allowed).

7. In the hyperbolic plane with let Pn be a geodesic n-polygonwhose n � 3 vertices are on the rim of H—hence not actually in H (Fig. 7.27).Find the area of Pn. (b) Deduce that the area of H(r) is infinite.

8. Stand the double torus M of Fig. 7.26 on end, and at each point, let Vbe the component of Uz = (0, 0, 1) that is tangent to M.

(a) Determine the indices of the singularities of V, and check that theirsum is c(M).(b) Do the same for a vector field whose integral curves are the level curvesz = const in M.

9. Use notation as in the definition of index of a vector field at an isolatedsingularity p (Def. 7.7), so j = –(Xa, Va) on [a, b].

(a) If ||Xa|| = 1, show that

where f = ·Va , XaÒ, g = ·Va, J(Xa)Ò.(b) Deduce that the index of V at p is the winding number of the planecurve (f, g): [a, b] Æ R2 - 0 (Ex. 5 of Sec. 4.6).

10. Referring to Fig. 7.25: (a) Sketch some integral curves of the meetingof three streams and find the index of the central singularity. What is theindex for the meeting of n > 3 streams?

j jb afg gff g

dta

b

( ) - ( ) =¢ - ¢

+Ú 2 2,

K r= -1 2/ ,


FIG. 7.27

(b) Do the same for the singularity with three leaves. What is the index forn > 3 leaves?(c) Deduce that there exist isolated singularities of every integer index.

11. For the vector field V = -uU1 + vU2 on R2,(a) Solve the differential equations u¢ = -u, v¢ = v explicitly to find the inte-gral curve of V starting at an arbitrary point (a, b) in R2. (For integralcurves, see Ex. 13 of Sec. 4.8.)(b) Sketch sufficiently many of these integral curves to decide what theindex of V at (0, 0) is.(c) Verify the index in (b) by using the integral in Exercise 9 above. (Hint:Take X = (1, 0).)

12. (Continuation by computer.) For the vector field V = (2u2 - v2)U1 + 3uvU2

on R2,(a) Numerically solve the differential equations for the integral curves ofV, and plot sufficiently many to decide what the index of V is at (0, 0).(Hint: Try the integral curve starting at the point (0, 1). See the Appendixfor numerical solution of differential equations.)(b) Verify the index in part (a) by a numerical integration derived fromExercise 9.

13. (Computer.) Same as the preceding exercise but with one sign changed:V = (2u2 - v2)U1 - 3uvU2.

7.8 Summary

A geometric surface—that is, a 2-dimensional Riemannian manifold—con-sists of an abstract surface furnished with an inner product on each tangentspace. The simplest case is a surface in R3, using the dot product of R3. Manygeometric features of surfaces in R3 survive unchanged in the more generalsetting (e.g., length of a curve, area of a surface). Others must have theirearlier definitions modified (e.g., geodesics, Gaussian curvature K). Stillothers, that involve R3 in an essential way, cannot be generalized at all (e.g.,mean curvature H).

The new definitions are not revolutionary. They appear naturally whenvectors normal to M Ã R3 are ignored. Previously, a geodesic had accelera-tion normal to M, so now it has acceleration zero. When the normal vectorE3 is dropped from an adapted frame E1, E2, E3, the connection forms reduceto a single 1-form w12, and the elegant formula dw12 = -K dM serves to defineGaussian curvature K.


The Gaussian curvature of a surface is its dominant geometric property,and as we have seen, curvature enters into almost every geometrical investi-gation. But its deepest consequence is the link between geometry and topol-ogy established by the Gauss-Bonnet theorem: The curvature of a compactsurface completely determines its topological structure.

7.8 Summary 387

▼

▲Chapter 8

Global Structure of Surfaces

388

In this chapter we investigate the global structure of geometric surfaces, thatis, 2-dimensional Riemannian manifolds. We want to know what the possi-ble surfaces are and what they are like. In maximum generality this goal isunrealistic, but under reasonable hypotheses, good results can be obtained.

The central theme of this chapter is the influence of Gaussian curvature ongeodesics. A significant preliminary result is that in any surface, the geome-try of a neighborhood of a point is completely described by curvature andthe geodesics radiating out from that point.

This local result can be extended to show that geodesics can grip an entiresurface. The first step is to show that geodesics starting at any point p ofa complete surface eventually reach every point of that surface (Section 2).Gaussian curvature controls the spreading and contracting of these geodes-ics as they radiate out to cover the surface, but their global pattern can bequite complicated.

Nevertheless, by using topological methods (Section 4), considerable globalinformation can be derived from this pattern. In particular, we give detailedresults in two broad cases: surfaces with constant curvature and surfaceswhose curvature obeys either K � 0 or K � k > 0.

8.1 Length-Minimizing Properties of Geodesics

The preceding chapters viewed geodesics as straightest curves (no turning);now we examine their character as shortest curves. For the Euclidean plane,the general problem of shortest routes is simple: Given any two points p andq in R2, there is a unique straight-line segment (geodesic) from p to q, andthis is shorter than any other curve from p to q.

For an arbitrary geometric surface the situation is more interesting. In thefirst place, there may be no shortest curve from p to q (Exercise 7 in Section6.4). And even if there is one, it may not be unique. For example, we willsoon prove the expected result that on the sphere, every semicircle from, say,the north pole to the south pole is shortest.

1.1 Definition Let a be a curve segment from p to q in M. Then(1) a is a shortest curve segment from p to q provided that if b is any other

curve segment from p to q, then

(2) a is the shortest curve segment from p to q provided it is a shortestcurve and any other shortest curve segment from p to q is a reparametriza-tion of a.

In case (1) we also say that a minimizes arc length from p to q. In terms ofintrinsic distance r in M (Definition 4.1 in Chapter 6), an equivalent defini-tion is L(a) = r(p, q).

In case (2) we say that a uniquely minimizes arc length from p to q. Unique-ness must be interpreted liberally here, since monotone reparametrizationdoes not change arc length (Exercise 7 in Section 2.2).

All such shortest curves will turn out to be geodesics, and our goal in thissection is to show that “short” geodesic segments in an arbitrary geometricsurface M behave as well as geodesics in R2. To do so we use a remarkablemapping that compares the region around any point p in M with the Euclid-ean plane. The tangent plane Tp(M) at p provides the plane, and the mappingis defined as follows, with gv, as usual, denoting the geodesic of M whoseinitial velocity is v.

1.2 Definition For a point p of a geometric surface M, the exponentialmap expp is given by

for all v in Tp(M) such that gv is defined on the interval [0, 1].

The exponential map is differentiable and is defined at least on a neigh-borhood of 0 in Tp(M)—a consequence of the fact that solutions of ordinarydifferential equations depend differentiably on initial conditions as well as onthe parameter.

The geometrical meaning of expp is clarified by

exp p vv( ) = ( )g 1

L Lb a( ) ( )� .

8.1 Length-Minimizing Properties of Geodesics 389

390 8. Global Structure of Surfaces

1.3 Lemma The exponential map expp: Tp(M) Æ M carries radial linesfrom 0 in Tp(M) to geodesics starting at p in M. Explicitly,

for all t such that expp(tv) is well-defined (Fig. 8.1).

Proof. For fixed t in R and v in Tp, the geodesic s Æ gv(ts) has initial veloc-ity tg ¢v(0). So does the geodesic gtv. Hence by the uniqueness of geodesics(Theorem 4.3 of Chapter 7), these geodesics are equal, that is, gv(ts) = gtv(s)whenever both sides make sense. In particular, setting s = 1 gives

◆

Thus the exponential map at p collects all the geodesics starting at p intoa single mapping. The best case is when the exponential map is defined onthe whole tangent space, as discussed in the next section. But in any case, theexponential map expp is well behaved near p.

1.4 Lemma For each p in M the exponential map at p carries some neigh-borhood U of 0 in Tp(M) diffeomorphically onto a neighborhood N of p inM (Fig. 8.1).

Here N is called a normal neighborhood of p. When U is an open disk ||v|| < e, the normal neighborhood is written as Ne and is said to have radius e.

Proof. (For the first time we treat the tangent plane as a surface in itsown right—and work with vectors tangent to it.) In view of the inversefunction theorem, it will suffice to show that the tangent map of expp at 0is a linear isomorphism, for then expp will be diffeomorphism on someneighborhood of 0.

g gv tv pt t( ) = ( ) = ( )1 exp .v

exp p vt tv( ) = ( )g

FIG. 8.1


A tangent vector v0 to Tp(M) at 0 is the initial velocity of the ray r(t) = tv, and we saw above that

Since tangent maps preserve velocities,

Thus the tangent map of expp at 0 is just the natural isomorphism v0 Æ v. ◆

Note that for a given point p there is always a largest—possibly infinite—normal radius e.

Now we apply the exponential map to the problem at hand. In the Euclid-ean plane, if one is interested in distance to the origin, it is natural to usepolar coordinates, since the distance from (0, 0) to (ucosv, usinv) is simplyu. Polar coordinates can be installed in an arbitrary geometric surface M withany point p as the pole. The first step is to put these coordinates into thetangent plane at p by choosing a frame e1, e2 at p and defining

Then the exponential map at p moves this parametrization into M.

1.5 Definition (1) Let e1, e2 be a frame at a point p of M. The mapping

is called a geodesic polar mapping.(2) If Ne is a normal e-neighborhood of p in M, and the mapping x(u, v)

is defined only for 0 � u < e, then x is called a geodesic polar parametrizationof Ne with pole p (Fig. 8.2).

x x e eu v u v u v u vp p, ,( ) = ( )( ) = +( )exp ˜ exp cos sin1 2

˜ cos sin .x e eu v u v u v u,( ) = + ( )1 2 0 � � e

exp * exp * .p p vv v0 0 0( ) = ¢( )( ) = ¢( ) =r g

exp .p vt tr g( )( ) = ( )

FIG. 8.2

Even though the full mapping x in (1) is usually neither one-to-one norregular, we will be able to use it along particular radial geodesics. Restrictedin (2) to a normal e-neighborhood, x is strictly speaking only a parame-trization of the punctured neighborhood Ne - p. However, the ambiguitieswhen u = 0 are familiar from Euclidean polar coordinates—and we need u = 0.

The u-parameter curve v = v0 of x is the radial geodesic with initial veloc-ity v = cosv0e1 + sinv0e2 since, using Lemma 1.3,

Since ||v|| = 1, this geodesic has unit speed, so its length from p = x(u, v) is just u.

The v-parameter curve u = u0 > 0,

parametrizes a closed curve called the polar circle of radius u0 and pole p (seeFig. 8.2).

Note that if q = x(u0, v0) is any point of Ne except p, then (but for repa-rametrization) there is only one unit-speed geodesic from p to q that liesentirely in Ne, namely, the radial geodesic

1.6 Lemma For a geodesic polar parametrization,

Proof. Since the u-parameter curves are unit-speed geodesics,

Thus

Thus F is constant on each u-parameter curve.The v-parameter curve v Æ x(0, v) is the constant curve at p, so

xv(0, v) = 0 for all v. This means that F(0, v) = 0 for all v. Since Fu = 0,we conclude that F is identically zero.

The results so far—that E = 1, F = 0—are valid if x is merely a geo-desic polar mapping. But now we assume that x is a parametrization of ane-neighborhood. Thus, restricted to 0 < u < e, x is a regular mapping, so

F Eu u v u u vu u uv v= = = = =x x x x x x, , , .12

0

E u u uu= = =x x x, .1 0,

E F G= = ( ) >1 0 0, , and except at the pole .

g u u v u u( ) = ( ) ( )x , 0 00 � � .

x e eu v u v u vp0 0 1 0 2, ,( ) = +( )exp cos sin

x e eu v u v u v up v, 0 0 1 0 2( ) = +( ) = ( )exp cos sin .g



◆

Here E = 1 means that the exponential map preserves radial distances,and F = 0 means that polar circles are always orthogonal to radial geodes-ics (Fig. 8.2). Only G can be non-Euclidean, that is, π 1. So a normal neigh-borhood in any surface M can be manufactured from a (flat) neighborhoodof 0 in R2 ª Tp(M) merely by stretching the polar circles u = const.

1.7 Example We explicitly work out geodesical polar parametrizationsin two classic cases.

(1) The unit sphere S in R3. For simplicity, we take the pole p to be thenorth pole (0, 0, 1). (By a scale change and a Euclidean rotation of S, essen-tially the same results will hold for any point in a sphere of any radius.) Toget geodesics radiating out from p, we change the geographical parametriza-tion to

Each u-parameter curve is a unit-speed parametrization of a circle of longi-tude through p, and hence is geodesic. The initial velocity of these radial geo-desics is

So for the frame e1 = U1, e2 = U2 (Fig. 8.3), the uniqueness of geodesicsimplies that

Hence x is indeed a polar geodesic parametrization. It is clear that the largestnormal neighborhood of p has radius p and fills all of S except the southpole (0, 0, -1). The radial geodesics are longitudinal semicircles, and the polarcircles are the circles of latitude.

x e ee eu v u u v u vv v p,( ) = ( ) = +( )+g cos sin exp cos sin .1 2 1 2

xu v v v vU vU0 0 1 2, , ,( ) = ( ) = +cos sin cos sin .

x u v u v u v u, , ,( ) = ( )sin cos sin sin cos .

G EG F= - >2 0.

FIG. 8.3

(2) The hyperbolic plane H (Example 2.5 of Chapter 7). We take the poleto be the origin 0 and let e1, e2 be the natural frame U1, U2. (Since h(0, 0) =1, this Euclidean frame is also a hyperbolic frame.) According to Example5.6 of Chapter 7, the geodesics of H through the origin follow Euclideanstraight lines. Thus for any number v, the curve

is at least a pregeodesic, and we found the arc length function of a to be

Thus, shifting notation from s to u, all the radial unit-speed geodesics can becollected in the mapping

Since

it follows, as in (1), that x is a geodesic polar parametrization. The largestnormal neighborhood in this case is the entire surface H. The radial geodes-ics are Euclidean lines, and polar circles are the Euclidean circles with center0 (both with non-Euclidean metric properties). ◆

1.8 Theorem For each point q of a normal neighborhood Ne of p, theradial geodesic segment of Ne from p to q uniquely minimizes arc length.

Proof. Let x be a polar parametrization of Ne. If q = x(u0, v0), then—admitting the value 0 for u—the radial segment out to q is

Now let a be an arbitrary curve segment from p to q in M. We param-etrize a on the same interval as g without changing its arc length. For g tobe a shortest curve from p to q, we must prove

Consider first the case where a stays in the neighborhood Ne (Fig. 8.4).We can assume that once having left p, a never returns—for if it

did, throwing away the resulting loop would shorten a. Thus it is possibleto write

L Lg a( ) ( )� .

g u u v u u( ) = ( ) ( )x , 0 00 � � .

x e eu v v v0 1 2, ,( ) = +cos sin

x u vu

v v, ,( ) = ( )22

tanh cos sin .

s tt( ) = Ê

Ëˆ¯

-22

1tanh .

a t t v t v( ) = ( )cos sin,



Since a(0) = p and a(u0) = q, we have

(1)

Because x has E = 1 and F = 0, the speed of a is .Now,

(2)

Hence

(3)

where the last step uses (1). Since the radial geodesic has unit speed,

and so we conclude that L(g ) � L(a).If a does not stay in Ne, then strict inequality, L(g ) < L(a), will hold.

In fact, to leave Ne, a must cross the polar circle u = u0; so it already haslength u0 = L(g )—and it has further to go.

Now we prove the uniqueness assertion:

The argument above shows that if L(a) = L(g ), then a stays inside Ne andthe inequality in (3) becomes an equality. The latter implies

If , then is a reparametrization ofL La g a g( ) = ( ) .

L du uu

g( ) = =Ú00

0

,

L a G a du a du

a u a u

u u

a( ) = ¢( ) + ¢( ) ¢

= ( ) - ( ) =

Ú Ú12

12

0

0

10

0

1 0 1 00

�

,

¢ + ¢ ¢ = ¢ ¢a Ga a a a12

22

12

1 1� � .

¢ = ¢ + ¢a a Ga12

22

a a u u

a v a u v1 1 0 0

2 0 2 0 0

0 0

0

( ) = ( ) =( ) = ( ) =

, ,

, .

a u a u a u u u( ) = ( ) ( )( ) ( )x 1 2 00, � � .

FIG. 8.4


Since G > 0, we conclude from (2) that

Thus a2 has constant value v0, so

This expresses a as a monotone reparametrization of g. ◆

This fundamental result shows, as promised earlier, that if points p, q in asurface M are close enough together, then—as in Euclidean space for arbi-trary points—there is a unique geodesic segment from p to q that is shorterthan any other curve in M from p to q.

1.9 Example Length-minimizing geodesics on the sphere. Let S Ã R3 be asphere of radius r. It follows from Example 1.7(1) that for each point p of Sthe largest normal neighborhood has radius pr and covers the entire sphereexcept for the point -p antipodal to the pole p. Hence the preceding theoremimplies:

(a) If distinct points p and q are not antipodal, q π ±p, then the radialgeodesic g from p to q is the unique shortest curve joining p and q. But weknow all the geodesics of S, so g can only be the one that follows the shorterarc of the great circle through p and q.

(b) Intrinsic distance on S is given by the formula

where J is the angle from p to q in R3 (Fig. 8.5). If q π ±p, this follows from(a), since

r g Jp q,( ) = ( ) =L r .

r J J pp q, ,( ) = ( )r 0 � �

a gu a u v a u( ) = ( )( ) = ( )( )x 1 0 1, .

¢ ¢ =a a1 20 0� and .

¢ + ¢ = ¢a Ga a12

22

1.

FIG. 8.5

Otherwise, r(p, p) = 0, and as q moves toward the antipodal point -p of p,continuity implies r(p, -p) = pr. Consequently,

(c) There are infinitely many minimizing geodesics from any point p to itsantipodal point -p, namely, constant speed parametrizations of the semicir-cles running from p to -p. (Proof. These all have length pr = r(p, -p).)

(d) No geodesic g of length L(g ) > pr can minimize arc length between itsend points. This follows immediately from the fact that intrinsic distance onS never exceeds pr. It is clear geometrically, since if g starts at p, its lengthexceeds pr as soon as it passes the antipodal point -p. But then the other arcof the same great circle is shorter than g. ◆

Although the geodesic structure of the hyperbolic plane is simpler thanthat of the sphere, Theorem 1.8 is still informative. Since the entire hyper-bolic plane is a normal neighborhood of the point 0, it tells us that the intrin-sic distance from 0 to any point p is the length of the radial geodesics from0 to p. The preceding example gives this explicitly as

where ||p|| is the Euclidean distance from 0 to p. It follows that every geodesicg of H has infinite length—hence H is complete. In fact, using the triangleinequality,

Then as t Æ •, r(0, g (t)) Æ •; hence L(g |[t0, t]) Æ •.The differential equations result used to establish the domain and differ-

entiability of the exponential map will, in fact, yield this local uniformityproperty of normal radii: For each point p of a surface M, there is an e > 0such that every point q with r(p, q) < e has a normal neighborhood of radiuse. (For a proof, see [doC], for example.)

Using this property, we can confirm the maxim that a shortest road has noturning.

1.10 Corollary If a is a shortest piecewise regular curve in M from p toq, then a is an (unbroken) geodesic.

Proof. First we show that the regular segments of a are geodesics; thenthat a has no corners. We can suppose a has unit speed.

Assume that a regular segment ai|[bi, bi+1] of a is not a geodesic. Thena ≤i (t0) is nonzero for some t0 in [bi, bi+1], and by continuity we can supposet0 < bi+1. Then a subsegment ai|[t0, t0 + e] is contained in a normal e-

L t t t t t tg r g g r g r g0 0 0, .[ ]( ) ( ) ( )( ) ( )( ) - ( )( )� �, , ,0 0

r 0 pp

, ,( ) = -22

1tanh



neighborhood of ai(t0). Note that ai|[t0, t0 + e] cannot be reparametrizedto be geodesic. Thus by the uniqueness feature of Theorem 1.8, the radialgeodesic segment s from ai(t0) to ai(t0 + e) is strictly shorter (Fig. 8.6).Replacing ai|[t0, t0 + e] by s changes a to a strictly shorter curve from pto q, contradicting the assumption that a is shortest. Thus a is a possiblybroken geodesic.

Now we assume that a actually has a corner, say at ai-1(bi) = ai(bi), andagain deduce a contradiction. By the remark preceding this lemma, thereis an e > 0 such that ai-1(bi) is contained in in a normal neighborhood N

of ai-1(bi - e). By continuity, some initial subsegment ai|[bi, t1] of ai is stillin N.

Thus the combined curve from ai-1(bi - e) to ai(t1) has a corner and liesin N. So it is strictly longer than the radial geodesic t of N joining thesepoints (Fig. 8.6). Then as before, replacing the combined curve by t short-ens a, giving the required contradiction. ◆

Exercises

1. Prove:(a) A normal e-neighborhood of p in M consists of all points q in M suchthat r(p, q) < e.(b) If p π q in M, then r(p, q) > 0. (This completes the proof that intrin-sic distance r is a metric on M; see Ex. 3 of Sec. 6.4.)

2. (Normal coordinates.) Let N = expp(U ) be a normal neighborhood of apoint p in M, and let e1, e2 be a frame at p. Prove:

(a) The mapping

is a coordinate patch on N.

(b) At p (but generally not elsewhere) E = 1, F = 0, G = 1. Thus normalcoordinates are Euclidean at p, hence at least approximately Euclidean near p.(c) Coordinate straight lines through p are geodesics of M.

n e ex y x yp,( ) = +( )exp 1 2

FIG. 8.6

(d) With suitable choices, n for R2 is the identity map n(x, y) = (x, y). Sofor arbitrary M, normal coordinates generalize the natural (rectangular)coordinates of R2.

3. At the point p = (r, 0, 0) of the cylinder M: x2 + y2 = r2, let e1 = (0, 1,0) and e2 = (0, 0, 1). Find an explicit formula for the normal parametriza-tion in Exercise 2. What is the largest normal neighborhood of the point p?

4. (Continuation.) Prove:(a) A geodesic starting at an arbitrary point p = (a, b, c) in the cylinder M does not minimize arc length after it passes through the antipodal linet Æ (-a, -b, t). (Only vertical geodesics through p fail to meet this line.)(b) If q is not on the antipodal line of p, there is a unique shortest geo-desic from p to q.(c) Derive a formula for intrinsic distance on the cylinder.

5. Let M be an augmented surface of revolution (Ex. 12 of Sec. 4.1). Prove,without computation:

(a) If M has only one intercept p on the axis of revolution, then every geo-desic segment g starting at p uniquely minimizes arc length.(b) If M has a second intercept q, then the conclusion in (a) holds if andonly if g does not reach q.

6. In M let a be a curve segment in M from p to q, and b a curve segmentfrom q to r. Joining a and b does not usually produce a differentiable curvefrom p to r since a + b will usually have a corner at q.

In this case, prove that there is a piecewise smooth curve g from p to r thatis arbitrarily close to a + b but strictly shorter: L(g ) < L(a) + L(b). (Hint:See proof of Cor. 1.10.)

Techniques from advanced calculus show that the corner can actually besmoothed away, leaving g differentiable throughout.

7. (Intrinsic distance is continuous.)(a) For p0 in M, show that the real-valued function p Æ r(p0, p) is con-tinuous; in fact, if r(p, q) < e, then |r(p0, p) - r(p0, q)| < e.(b) State precisely and prove that the function (p, q) Æ r(p, q) is continuous.

8. The radial geodesics from the point (0, 1) in the Poincaré half-plane Pare given by (x - a)2 + y2 = a2 + 1 for all a (see Ex. 1 of Sec. 7.5).

(a) Show that the curves given by x2 + (y - b)2 = b2 - 1 for all b > 1 areeverywhere orthogonal to these geodesics.(b) Deduce that the curves in (a) are the polar circles about the pole (0, 1).(c) (Computer graphics.) Plot a few curves from each family on the samefigure.


8.2 Complete Surfaces

A geometric surface M is complete provided all its geodesics can be definedon the entire real line (Definition 4.4 of Chapter 7). Thus the plane R2 is com-plete, but any smaller open set in R2 is not. Completeness is a very naturalcondition. Physically, we can picture a point in a frictionless surface M as apenny p, constrained by a normal force to remain in M. Once p is given aninitial velocity, its motion is completely determined, and since there is noacceleration, it traces out a geodesic of M. It is reasonable to think that thismotion can be stopped only by a flaw in the surface.

In this section we consider some consequences of completeness. The mostremarkable is that the geodesics of a complete surface, assumed only to runforever, actually go everywhere—and in the shortest possible arc length.

2.1 Theorem (Hopf-Rinow) Any two points p and q in a complete con-nected geometric surface M can be joined by a shortest geodesic segment.

Proof. The scheme is an ingenious one, which begins by picking apromising candidate for the shortest curve (see Theorem 10.9 of [Mi]). Let

parametrize a polar circle Ca of radius a in a normal neighborhood of p. ByExercise 7 of the preceding section, the function v Æ r(b(v), q) is continuouson the closed interval [0, 2p], so it takes on a minimum value at some v0. Thusb(v0) is the point of Ca nearest to q. This makes it reasonable to hope that theradial geodesic through b(v0), namely

will hit target point q.Since M is complete, g is defined for all u � 0. We will, in fact, show

(1)

(See Fig. 8.7.) Furthermore, since g has unit speed, (1) implies

so g minimizes arc length as required.

To prove (1) we will show that

(2)r g u r u a u r( )( ) = -, for allq � � .

L rg r( ) = = ( )p q, ,

g rr r( ) = = ( )q p q, where , .

g u u v( ) = ( )x , ,0

b pv a v v( ) = ( ) ( )x , 0 2� �


8.2 Complete Surfaces 401

This is a kind of efficiency condition on g : Each inch that g advances bringsit one inch closer to q. Note that (2) does imply (1), since for u = r it givesr(g (r), q) = 0, hence g (r) = q.

We separate the proof of (2) into three parts.Start. Equation (2) holds for u = a (the radius of Ca); that is,

(3)

To show this, note that by Theorem 1.8, r(p, g (a)) = a. Hence by the triangle inequality,

To get (3) we must reverse this inequality. By the definition of intrinsic dis-tance, for any e > 0 there is a curve segment a from p to q such that

Now a must hit the polar circle Ca, say at a(t0), and we observe that thepart of a from p to a(t0) has length L1 � a, while the remainder has length

(The latter inequality holds since g (a) is a nearest point to q on C.) Thus

Since e was arbitrary, we get the inequality,

required to prove (3).Plan. The set of numbers u in [a, r] for which (2) holds has a least upper

bound b � r. Since the functions involved in (2) are continuous, it followsthat (2) holds for u = b, that is, r(g (b), q) = r - b. Evidently it suffices toshow that b = r.

a a+ ( )( ) ( )r g r, , ,q p q�

a a L L+ ( )( ) + ( ) +r g r e, ,q p q� �1 2 .

L t a2 0� �r a r g( )( ) ( )( ), ,q q .

L a r e( ) ( ) +� p q, .

r a a= ( ) + ( )( )r r gp q q, � , .

r g a r a( )( ) = -, q .

FIG. 8.7


To do this, our plan is to assume b < r and deduce a contradiction by proving

Indeed, this contradicts the least upper bound definition of b.Finish. Let C* be a polar circle of radius 0 < a* < r - b in a normal

neighborhood of g (b). By reproducing the argument for the circle Ca, weget a point c* such that

(See Fig 8.8.) Since r(g (b), q) = r - b, this becomes

(4)

It remains only to show that

(5)

By the triangle inequality,

Then (4) implies

But there is a broken geodesic from p to c* whose length is b + a*. In fact,as Fig. 8.8 shows, we can travel from p to g (b) with arc length b, and thenfrom g (b) to c* on a radial geodesic of length a*. Thus by Corollary 1.10,this curve is not broken. Hence it is g all the way, which means that g (b + a*) is exactly c*.

Finally, we substitute (5) into (4) to get the required contradiction

◆

Here are two fundamental consequences of completeness.

r g b a r b a+( )( ) = - +( )* * ., q

r p c, * *.( ) +� b a

r r rp c c q p q, , ,* * .( ) + ( ) ( ) =� r

c* * .= +( )g b a

r c q*, * .( ) = - +( )r b a

r r gc q q* *., ,( ) = ( )( ) -b a

r g b a r b a a+( )( ) = - +( ) >* * * ., for some numberq 0

FIG. 8.8

2.2 Corollary At every point p of a complete connected surface M theexponential map expp is defined on the entire tangent space Tp(M) and mapsit onto M.

Proof. By completeness, if v is tangent to M at p, the geodesic gv isdefined on the whole real line. But gv(1) is the definition of expp(v), so expp

is defined on all of Tp(M).By the Hopf-Rinow theorem, for any other point q in M there is a

geodesic segment g from g (0) = p to g (r) = q. Thus expp(rg ¢(0)) =g (r) = q. ◆

A connected geometric surface M is said to be extendible if it is isometricto an open subset of a strictly larger connected surface . Thus, for example,R2 - 0 is extendible to R2, but the following result shows that R2 itself is inextendible.

2.3 Theorem A complete connected surface M is inextendible.

Proof. We assume M is extendible and deduce a contradiction. For sim-plicity, let M actually be a subset of . Then it is an open set of (Exer-cise 15 of Section 4.3). It will suffice to find a geodesic g of that meetsM. For then the portion of g in M is a geodesic of M that cannot beextended—in M—over the whole real line. But this contradicts the com-pleteness of M.

Since is connected, the open set criterion (Exercise 9 of Section 4.7)implies that - M is not open. Hence there is a point p in - M suchthat every neighborhood N of p meets M. When N is a normal neigh-borhood, some radial geodesic from p will meet M and is thus the requiredgeodesic g. ◆

Completeness, a geometric condition, is implied by the topological prop-erty compactness.

2.4 Corollary A compact geometric surface is complete.

Proof. If every unit-speed geodesic in M can be extended by some fixedamount h > 0, then M is complete, because these small extensions can berepeated indefinitely.

By the remark preceding Corollary 1.10, for each point p in M there isan e > 0 such that every point q within distance e of p has a normal neigh-borhood of radius e. Since M is compact, a finite number of such neigh-borhoods, say, N1, . . . , Nk, cover all of M. Let 2h be the smallest of thecorresponding radii e1, . . . , ek.

MMM

MMM

M

8.2 Complete Surfaces 403

Now we can extend any geodesic by h. Consider g : [0, b) Æ M. Fromg (b - h) there will be radial geodesics of length 2h in all directions. Theone that has initial velocity g ¢(b - h) will then run for arc length 2h andhence extend g by h. ◆

In fact, a more elaborate proof shows that every closed surface M in R3 iscomplete. Although it applies only to surfaces in R3, this result is strongerthere than the corollary above. For example, it tells us that any quadricsurface in R3 is complete.

The properties above show that completeness is a crucial prerequisite forthe global study of surfaces.

Exercises

1. Show that the converse of the Hopf-Rinow theorem is false: Give anexample of a geometric surface M such that any two points can be joined bya minimizing geodesic segment but M is not complete.

2. Test the scheme used to prove the Hopf-Rinow theorem (2.1) as follows.Given points p = (p1, p2) and q = (q1, q2) in M = R2, use only that schemeto find a formula for a geodesic g starting at p and aimed at q.

3. Prove that every complete generalized cylinder C is isometric to the planeR2 or to a circular cylinder with radius uniquely determined by C.

4. Let M Ã R3 be flat. Exercise 2 of Section 6.3 correctly suggests that everypoint of M has a neighborhood that is ruled. Give an example to show thatM itself need not be ruled. (Hint: Begin by cutting a small square from eachcorner of a plane square.)

However, it is known that every complete flat surface in R3 is a generalizedcylinder (so Ex. 3 applies). See W. S. Massey, “Surfaces of Gaussian Cur-vature Zero in Euclidean Space,” Tohoku Math. J. 14, 1962.

5. Many surfaces of revolution with constant curvature K π 0 were foundin Section 7 of Chapter 5 and its exercises. Show that except for the sphere,none are complete.

6. Give an example of a surface in which every pair of points can be joinedby a geodesic segment, but not always by a minimizing one.

7. Let C be the z > 0 part of the cone z2 = x2 + y2 in R3. Show that C isnot complete, but that any two points of C can be joined by a minimizinggeodesic. (Hint: Cut C and bend it isometrically into R2.)


8.3 Curvature and Conjugate Points

The preceding section dealt with curve segments g from p to q whose arclength L is a minimum compared to all curve segments joining these points.Now we consider g on which L has only a local minimum. The definition hasthe same pattern as for a local minimum of a real valued function on R, butwith points of R replaced by curve segments from p to q.

3.1 Definition A geodesic segment g from p to q in M locally minimizesarc length from p to q provided that L(a) � L(g ) holds for every curvesegment a from p to q that is sufficiently close to g.

To clarify the term “sufficiently close,” we define a to be e-close to g pro-vided there is a reparametrization of a defined on the same interval I as gand such that

Then the ending of the preceding definition becomes:

provided there exists an e > 0 such that L(a) � L(g ) holds for anycurve segment from p to q that is e-close to g.

The local minimization is strict (or unique) provided strict inequality L(a) > L(g ) holds except when a is a reparametrization of g.

To get an intuitive picture of this definition, we can imagine that g is anelastic string—or rubber band—that (1) is constrained to lie in M, (2) is undertension, and (3) has its end points pinned down at p and q.

Because g is a geodesic, it is in equilibrium. If it were not a geodesic, itstension would pull it to a new shorter position. But is the equilibrium stable?That is, if g is pulled aside slightly to a new curve a and released, will it returnto its original position? (See Fig. 8.9.) Evidently g is strictly stable if and onlyif g is a strict local minimum in the sense above, for if a is always longer thang, its tension will pull it back to g.

The key to local minimization is the notion of conjugate point. If g is aunit-speed geodesic starting at p, then g is a u-parameter curve v = v0 of ageodesic polar mapping x with pole p. We know that along g the function G = ·xv, xvÒ is zero at u = 0, but is nonzero immediately thereafter (Lemma1.6). A point g (s) = x(s, v0) with s > 0 is called a conjugate point of g (0) = pon g provided G(s, v0) = 0. (Such points may or may not exist.)

The geometric meaning of conjugacy rests on viewing as the dis-tance between nearby radial geodesics and hence of its derivative as( )G u

G v= x

r a g e˜ .t t t I( ) ( )( ) <, for all in

a

8.3 Curvature and Conjugate Points 405


the rate at which the radial geodesics are spreading apart. In fact, the dis-tance from x(u, v) to x(u, v + e) is approximately e ||xv(u, v)||, and ifthis distance is increasing, so the radial geodesics are spreading apart, whileif the distance is decreasing, so the radial geodesics are pullingcloser together.

Since G vanishes at a conjugate point g (s1) = x(s1, v0), we can expect that,for v near v0, the radial geodesics gv will all reach this point at distance s = s1 (Fig. 8.10). Unfortunately, this meeting may not actually occur, sinceG controls only first derivative terms of x, and higher order terms may stillbe nonzero when G vanishes.

The Euclidean plane R2 sets the standard rate at which radial geodesicsspread apart, and for x(u, v) = (ucosv, usinv) we have

In particular, there are no conjugate points. Let us compare this with the casesin Example 1.7, the unit sphere S and the hyperbolic plane H.

G u G u= =, hence ( ) .1

( )G u < 0

( )G u > 0

FIG. 8.9

FIG. 8.10


For S, since xv = (-sinu sinv, sinu cosv, 0), we find

Thus radial geodesics from the north pole (meridians of longitude) begin inEuclidean fashion but as cosu drops below 1, they spread apart ever lessrapidly than in R2. After they pass the equator (at u = p/2) cosu turns nega-tive and the geodesics begin crowding closer together. All have their first

conjugate point after traveling distance p, since . In thiscase, of course, the meeting actually takes place—at the south pole of S (Fig.8.3).

For the hyperbolic plane, we know that radial geodesics from the originfollow Euclidean straight lines. From the formula for x(u, v) in Example1.7(2), we can compute†

Thus the radial geodesics, after beginning in Euclidean fashion, spread apartever more rapidly than in R2, as might be guessed from the slogan “rulersshrink as they approach the rim.” In particular, there are no conjugate points.

3.2 Theorem (Jacobi) If g is a geodesic segment from p to q such that thereare no conjugate points of p = g (0) on g, then g locally minimizes arc length(strictly) from p to q.

Proof. For a geodesic polar mapping x with pole p, we can write

Since there are no conjugate points of p on g, we have

As noted earlier, it follows from Lemma 1.6 that EG - F 2 reduces to Gfor a geodesic polar mapping, so x is regular for all (u, v0) with 0 < u � b.Thus the image of g for 0 < u < b is covered by neighborhoods N that arediffeomorphic images, under x, of open sets in the uv plane.

Then g can be divided into segments by numbers

so that the first segment g |[0, u1] lies in a normal neighborhood N0 of pand each later segment g |[ui, ui+1], 1 � i � k, is in one of the neighbor-hoods N, say Ni.

0 1 1< < < < =+u u u bk k. . . ,

G u v u b, for0 0 0( ) > < < .

g gu u v u b L( ) = ( ) = ( )x , for0 0 � � .

G u G uu= =sinh , ( ) cosh .hence

G vp p,( ) = =sin 0

G u G uu= =sin , ( ) cos .hence

† We will soon find a method much quicker than direct computation.


Now we can choose an e > 0 so small that if a curve a is e-close to g,then each subsegment a |[ui, ui+1] lies in the same open set Ni as g |[ui, ui+1]and e is less than the radius of the normal neighborhood N0 (Fig. 8.11).

The first segment of a can evidently be written in the polar form

a(t) = x(a1(t), a2(t)),

with a1(0) = 0, a2(0) = v0. The local inverses of x on each Ni carry the latersegments, in succession, back to the uv plane. Thus a can be written as

with

We must show that L(a) � L(g ), with equality only if a is a monotonereparametrization of g. As in the proof of Theorem 1.8, we find

And if L(a) = L(g ), then as before, a is a monotone reparametrization of g. ◆

The study of conjugate points can be radically simplified by freeing it fromdependence on geodesic polar mappings. To achieve this, we examine the

“spreading coefficient” more closely.

3.3 Theorem Let x be a geodesic polar mapping on whose domain in the uv plane, G > 0 if u > 0. Then satisfies the Jacobi differentialequation

( )G K Guu + = 0,

G v= x

( )G u

L a Ga a dt

a b a b L

b b

a

g

( ) = ¢ + ¢ ¢

= ( ) - ( ) = = ( )Ú Ú1

22

2

01

0

1 1 0

�

.

a a v a b b a b v1 2 0 1 2 00 0 0( ) = ( ) = ( ) = ( ) =, ,; .

a t a t a t( ) = ( ) ( )( )x 1 2, ,

FIG. 8.11

subject to the initial conditions

The restriction G > 0 is needed to ensure that is differentiable, at least foru > 0.

Proof. The Jacobi equation follows immediately from the curvatureformula in Proposition 6.3 of Chapter 6. As noted in the proof of Lemma1.6, x(0, v) = p for all v, so G(0, v) = 0 for all v. Because is notdifferentiable at u = 0, limits are required, and we must show

It is only necessary to consider a single radial geodesic g (u) = x(u, v0),setting

On g, since E = 1, F = 0, we get a frame field

Because g is a geodesic, E1 is parallel, and by Exercise 2 of Section 7.3,so is E2. By parallelism, E2 is well-defined at u = 0. Now,

Hence

Furthermore, since xv = gE2 on g, and E2 is parallel, we find

on g, for u > 0. Taking limits as u Æ 0 yields

(*)

But xu(0, v) = cosv e1 + sinv e2 for all v. Hence

Comparing this equation with (*) shows that limuÆ0 g¢(u) = 1. Since g = , this is the required limit. ◆G

x e euv v v v E0 00 0 1 0 2 2,( ) = - + = ( )sin cos .

xuvu

v g u E0 000

2, lim .( ) = ¢( ) ( )Æ

x xuv vu g E= = ¢ 2

E v v2 0 1 0 20( ) = - +sin cos .e e

E v v vu1 0 0 1 0 20 0( ) = ( ) = +x e e, cos sin .

E E gu v1 2= ¢ = =g x x, / .

g u G u v u( ) = ( ) >( ), 0 0 .

lim( ) .G u vu uÆ

( ) =0

1,

G v= x

G

G v G u v vu

u0 0 10

, , , for all( ) = ( ) =Æ

lim( ) .



Recall that for the Euclidean plane we found for u > 0, and hence. Thus the initial conditions in the preceding theorem show that as

radial geodesics first leave the pole p in any geometric surface, they arespreading at the same rate as in R2. Thereafter, the Jacobi equation

shows that the rate of spreading depends on Gaussian curvature. For K < 0,radial geodesics spread faster than in R2, as we saw earlier for the hyperbolicplane. For K > 0, they spread slower than in R2, as on the sphere.

To locate conjugate points it is no longer necessary to construct a geodesic polar mapping, as we have done so far. We can find on a geodesic gsimply by solving the Jacobi equation along g, subject to the Jacobi initialconditions. Explicitly, Theorem 3.3 gives

3.4 Corollary Let g be a unit-speed geodesic starting at p in M. Let g bethe unique solution of the Jacobi equation on g,

that satisfies the initial conditions g(0) = 0, g¢(0) = 1.Then the conjugate points of g (0) = p on g are the points g (s), s > 0, at

which g(s) = 0.

As in the plane R2, conjugate points may not exist, but if they do, the firstone is particularly important because of Theorem 3.2.

3.5 Example Conjugate Points.(1) Let g be a unit-speed geodesic starting at any point p of the sphere S

of radius r. Since K = 1/r2, the Jacobi equation for g is g≤ + g/r2 = 0, whichhas the general solution

The initial conditions g(0) = 0, g¢(0) = 1 then give

The first zero s1 > 0 of g occurs at distance s1 = pr. Thus the first conjugatepoint of p occurs at the antipodal point -p. This agrees with our earlier com-putation using geodesic polar parametrization.

g s rsr

( ) = ÊË

ˆ¯sin .

g s Asr

Bsr

( ) = ÊË

ˆ¯ + Ê

Ëˆ¯sin cos .

¢¢ + ( ) =g K gg 0,

G

( )G K Guu = -

( )G u = 1G u=


(2) On a torus of revolution with radii R, r > 0, let g be a unit-speed param-etrization of the outer equator. Now g is a geodesic, and along it K has con-stant value 1/(r(R + r)). Thus by the preceding corollary the first conjugatepoint of, say, g (0) will occur at exactly the same distance as if g were on asphere with this curvature, so

3.6 Corollary There are no conjugate points on any geodesic in a surfacewith curvature K � 0. Hence every geodesic segment in such a surface islocally minimizing.

Proof. Apply Corollary 3.4 to a geodesic in M. The initial conditionsg(0) = 0 and g¢(0) = 1 show that g starts out as a strictly increasing pos-itive function. Since K � 0, the Jacobi equation gives

Thus g¢(s) ≥ 1 for all s. Then evidently g(u) > 0 for all u > 0. ◆

For example, on a (flat) circular cylinder, the helical geodesic g from p toq indicated in Fig. 8.12 is indeed stable, as one can verify by experiment.Although locally minimizing, it is certainly not minimizing, since the straight-line segment s gives a much shorter way to travel from p to q.

Although geodesics locally minimize arc length before the first conjugatepoint, they do not locally minimize past it.

3.7 Theorem Let g be a geodesic segment from p to q. If there is a con-jugate point of p along g before q, then g does not locally minimize arc lengthbetween p and q.

¢¢ = -g Kg � 0.

s r R r1 = +( )p .

FIG. 8.12

A formal proof requires the calculus of variations (see [Mi] or [dC]), butwe can give a persuasive argument. As before, let g be the u-parameter curvev = v0 of a geodesic polar mapping x. If q = g (u0) = x(u, v0), then by hypoth-esis, the function u Æ G(u, v0) is zero at some number s with 0 < s < u0.

We have seen that this means that nearby u-parameter geodesic segmentsof the same length as g tend to meet again at g (s) = x(s, v0) (see Fig. 8.10).Suppose that (as on the sphere) this meeting actually occurs for some v1 arbi-trarily near v0. Then we construct the broken geodesic b : [0, u0] Æ M for which

Thus b has the same length as g |[0, u0]. But it has a corner at g (s), and as theproof of Corollary 1.10 shows, cutting across this corner produces a strictlyshorter curve that is arbitrarily near to b and hence to g. Thus g does notlocally minimize arc length from p to q.

Note that the proof of Corollary 1.10 remains valid for local minimiza-tion, so a locally minimizing curve from p to q is an (unbroken) geodesic.

In the critical case where the end point g (b) of a geodesic segment g : [a, b] Æ M is exactly the first conjugate point of its initial point g (a),nothing can be said in general: g may or may not locally minimize arc length.

The Jacobi differential equation can be used to give an intuitive descrip-tion of Gaussian curvature. First we need

3.8 Lemma If x is a geodesic polar mapping with pole p, then

Here o(u3) denotes a function of u > 0 and v such that limuÆ0o(u3)/u3 = 0.So in the formula, when u is small enough, o(u3) is negligible compared tothe first two terms.

Proof. Again consider on a radial geodesic u Æ x(u,v).As a solution of the Jacobi equation, g is differentiable at u = 0, and hencehas a Taylor expansion,

We can evaluate all these coefficients. The Jacobi initial conditions in Corollary 3.4 are g(0) = 0, g¢(0) = 1. Using the first of these in the Jacobi equation gives g≤(0) = 0. Now differentiate the Jacobi equation toget

g u g g u gu

gu

o u( ) = ( ) + ¢( ) + ¢¢( ) + ¢¢¢( ) + ( )0 0 02

06

2 33 .

g u G u v( ) = ( ),

G u v u Ku

o u u,( ) = - ( ) + ( ) ( )p3

3

60� .

bg

uu v u s

u s u u( ) =

( )( )

ÏÌÓ

x , if

if1

0

0 � �

� �

,

.


Thus

Substitution in the Taylor expansion then gives the required result. ◆

Suppose the inhabitants of a geometric surface M want to determine theGaussian curvature of M at a point p. By measuring a short distance e in alldirections from p, they obtain the polar circle Ce of radius e. If M = R2, thenCe is just an ordinary Euclidean circle, with circumference L(Ce) = 2pe. Butfor K > 0 the radial geodesics from p are not spreading as rapidly, so Ce shouldbe shorter than 2pe, and for K < 0 they are spreading more rapidly, so Ce

should be longer than 2pe.The dependence of L(Ce) on K can be measured with precision. For e small

enough, Ce is parametrized by v Æ x(e, v), where x is a geodesic polar patchat p. Thus

Hence by the preceding lemma,

(*)

Thus if surveyors in M measure L(Ce) carefully for e small, they can estimatethe Gaussian curvature of M at p to any desired accuracy. Taking limits yieldsa precise result.

3.9 Corollary .

Let us test the formula (*) on a sphere S of radius r in R3. As Fig. 8.13shows, the polar circle Ce with center p is actually a Euclidean circle of Euclid-ean radius r sinJ, where J = e/r. Then by the Taylor series of the sine function,

Comparison with (*) gives yet another proof that the sphere has Gaussiancurvature K = 1/r2.

L C rr r

os( ) = ÊË

ˆ¯ = - + ( )Ê

Ëˆ¯2 2

6

3

23p

ep e

eesin .

K L Cp( ) = - ( )( )Æ

lime

epepe

0 3

32

L C K os( ) = - ( ) + ( )ÊË

ˆ¯2

6

33p e

eep .

L C G v dve

p

e( ) = ( )Ú , .0

2

¢¢¢( ) = - ( )( ) = - ( )g K K0 0g p .

¢¢¢ + ( )¢ + ( ) ¢ =g K g K gg g 0.



Exercises

1. If x(u, v) is a polar parametrization in a surface of constant curvaturek, show that

2. In a normal neighborhood of p Œ M, call the region De on and withinthe polar circle Ce a polar disk of radius e.

(a) Show that the area of a polar disk is

and hence

(b) Use this formula to find the Gaussian curvature of a sphere ofradius r.

3. (a) At the pole 0 in the hyperbolic plane H, find the length of the polarcircle Ce and the area of the polar disk De, where 0 < e < • is the hyper-bolic radius.(b) Deduce from each that K(p) = -1.

(As we shall later see, these results hold for any point of H.)

KA D

p( ) =- ( )

Æ

120

2

4ppe

ee

elim .

A D K oe p ee

e( ) = - ( ) + ( )ÊË

ˆ¯

24

4

12p ,

G u v

rur

kr

u k

rur

kr

,

sin ,

sinh .

( ) =

=

= -

Ï

ÌÔÔ

ÓÔÔ

if

if = 0,

if

1

1

2

2

FIG. 8.13


4. Let M be an augmented surface of revolution (Ex. 12 of Sec. 4.1).(a) If M crosses the axis A at only one point p (as on a paraboloid of rev-olution), show that p has no conjugates on any geodesic.(b) If M crosses the axis A at only two points p and q (as on an ellipsoidof revolution), show that p and q are conjugate along every meridian.(Hint: Use a canonical parametrization to get the Jacobi equation.)

The following exercises deal with a useful variant of the geodesic polarparametrization in which the pole p is replaced by an arbitrary regular curve.

5. Let b: I Æ M be a regular curve in M, and let X be a nonvanishing vectorfield on b such that b¢(v) and X(v) are linearly independent for all v. Define

Thus the u-parameter curve v = v0 is a geodesic cutting across b with initialvelocity X(v) (Fig. 8.14). Prove:

(a) x is a regular mapping on some open region D in R2 containing thepoints (0, v) for all v in I.(b) By suitable choices of b and X, this parametrization x becomes (i) theidentity map of R2 (natural coordinates), (ii) the canonical parametriza-tion of a surface of revolution, and (iii) a ruled parametrization of a ruledsurface (Def. 2.6 in Ch. 4).

(Note: To obtain familiar formulas it may sometimes be necessary to reverseu and v.)

6. (Continuation.) If b is a unit-speed curve and X is the unit normal N = J(b¢) in M, show that E = 1, F = 0, and is the solution of the Jacobi

equation such that( )G K Guu + = 0

G

x u v u uX vX v v,( ) = ( ) = ( )( )( ) ( )g bexp .

FIG. 8.14


By analogy with conjugate points, if G(u0, v0) = 0, we say that x(u0, v0) is afocal point of b along the normal geodesic v = v0. Here light rays emergingorthogonally from b tend to meet (Fig. 8.15).

7. (a) If b is a circle of latitude on a sphere S, show that the north andsouth poles of S are the only focal points of b.(b) If b is a curve in the Euclidean plane, show that its focal points areexactly its centers of curvature, that is, the points on its evolute. (See Ex.13 of Sec. 2.4.)

8. (a) Let x and be geodesic polar parametrizations of normal e-neigh-borhoods Ne and e (same e) in two (not necessarily different) geomet-ric surfaces. If K(x) = K( ) on the common domain of x and , provethat Ne and e are isometric.(b) Deduce Minding’s theorem that constant curvature uniquely determineslocal geometry: If M and M¢ have the same constant curvature, then anypoints p in M and p¢ in M¢ have isometric neighborhoods.

8.4 Covering Surfaces

There is a close relationship between certain pairs of surfaces that lets infor-mation about one—obtained perhaps with difficulty—be transmitted easilyto the other. We are interested mostly in surfaces, but we state the basic def-inition for manifolds of arbitrary dimension so that Curves (1-dimensionalmanifolds) are included.

4.1 Definition Let F be a differentiable mapping of a manifold M ontoa manifold N of the same dimension. An open set V in N is evenly coveredprovided the set of points in M that map into V splits into disjoint open sets,each mapped diffeomorphically onto V by F. Then F is a covering map pro-vided every point of N has an evenly covered neighborhood.

N

xxN

x

G v G v vu g0 1 0, ( ) , .( ) = ( ) = - ( )and k

FIG. 8.15

8.4 Covering Surfaces 417

Here “diffeomorphically” means that the restriction of F to any one ofthese disjoint open sets, say U1, is a diffeomorphism of U1 onto V. Thus F |U1

has a differentiable inverse mapping l1: V Æ U1. Note that any open subsetof an evenly covered neighborhod is itself evenly covered.

If F: Æ M is a covering map, then is called a covering manifold ofM, and we say that covers M.

4.2 Example The classic 1-dimensional example of a covering map is theexponential map

which wraps the line R around the unit circle C: x2 + y2 = 1 in R2.† Figure 8.16 presents R as an infinite coil, with E dropping each point of Rdown to C; the evenly covered neighborhoods are readily pictured (see alsoExercise 4.2).

The exponential map E resolves the ambiguity between the “geometricangle” formed by two vectors (or two radial lines) and the “numerical angle”that measures it. Here a point (x, y) of C represents the geometric anglebetween the positive x axis and the line from the origin through the point (x, y), and J is an oriented angle (using the natural orientation of the plane).For any point (x, y) in C, the infinitely many numbers sent to (x, y) by E arethe possible measurements of the angle—any two differing by an integer multiple of 2p.

It is clear from the definition that a covering map Æ M is a local diffeomorphism onto M, but we will soon see that the converse is not true.

M

E eiJ J JJ( ) = = ( )cos sin, ,

MMM

FIG. 8.16

† This mapping, with R considered as a tangent line to C, was a model for the general notion of theexponential map in Section 1.

4.3 Example Several covering maps have appeared earlier.

(1) The usual parametrization x: R2 Æ T of the torus of revolution.(2) The map F(x, y) = (cosx, sinx, y) that wraps the Euclidean plane R2

around a cylinder C Ã R3.(3) The projection F of the sphere S onto the projective plane P

(Example 8.2 in Chapter 4).(4) The natural map of the orientation covering of a surface M onto

M (Exercise 6 of Section 4.8). ◆

A covering map Æ M has the crucial property that curves in M canalways be lifted to in the following sense.

4.4 Proposition Let F: Æ M be a covering map. If a: I Æ M is acurve and p is any point of such that F(p) = a(t0) in M, then there is aunique curve a : I Æ such that

Proof. For simplicity, take I = [0, b] and t0 = 0.Existence of Lifts. Since every point a(t) is in an evenly covered neigh-

borhood, a standard result from advanced calculus asserts that there arenumbers

such that for i = 1, … , n, each subinterval [ti-1, ti] is contained in an evenlycovered neighborhood, say Vi. We build on [0, b] by lifting a on eachsubinterval in succession.

There is no choice as to the lift of a |[0, t1]. Let U1 be the covering neigh-borhood of V1, that contains p. Since the covering neighborhoods are dis-joint, a lift of a |[0, t1] that starts at p must lie entirely in U1. But F carriesU1 diffeomorphically onto V1, so this lift can only be

where l1: V1 Æ U1, is the inverse diffeomorphism of F|V1.Now we repeat this process, replacing p by (t1) to get a curve 12 such

that F( 12) = a |[t1, t2]. This curve starts at the end point of a1, so the twocombine to give a curve 2 such that

After a finite number of such repetitions the entire curve is lifted.

˜ ˜ , .a a a2 2 20 0( ) = ( ) = [ ]p and F t

aa

aa

˜ : ã l a1 1 1 10= ( ) [ ] Æ, ,t M

a

0 0 1= < < < =t t t bn. . .

˜ ˜ .a a at F0( ) = ( ) =p,

MMM

MM

M


Uniqueness of Lifts. Here we only assume that F: Æ M is a local dif-feomorphism. Then if a1, a2: [0, b] Æ are lifts of the curve a in M suchthat a1(0) = a2(0), we must show that a1 = a2. The proof is based on thefact that [0, b] is not the union of two disjoint open sets.

Let A Ã [0, b] consist of all t such that a1(t) = a2(t). Since curves arecontinuous, the set [0, b] - A is open. But A is also open, since if t is in Athen there is a neighborhood U of a1(t) = a2(t) that is mapped diffeo-morphically onto a neighborhood V of a(t) in M. For t¢ near t, a(t¢) is inV, so, arguing as above, both a1(t¢) and a2(t¢) can only be l(a(t¢)); hence

a1(t¢) = a2(t¢).

Since A actually contains the number 0, we conclude that A is the entireinterval [0, b].

Note that the result holds if instead of a1(0) = a2(0), we assume thata1 and a2 agree at any t0 in [0, b]. ◆

In short, a curve in M can be uniquely lifted to any level in a coveringsurface of M.

Notation: For any map F, let F-1 (y) be the set of all x such that F(x) = y.

4.5 Corollary Let F: Æ M be a covering map with M connected, andlet k be a positive integer. If for some one q0 in M there are exactly k pointsin F-1(q0), then the same is true for every point q in M. In this case the covering is said to have multiplicity k.

Proof. For any q in M, let a: [0, 1] Æ M be a curve segment from q0 toq. Then let a1, . . . , ak be the lifts of a starting at the k points of F-1(q0).The end points i(1) all lie in F-1(q), and they are all different, for by theuniqueness of lifts, if any two end points agreed, the entire curves wouldbe identical, but by construction their initial points differ.

Consequently, F-1(q) contains at least as many points as F-1(q0). Clearlythe argument holds with q0 and q reversed, so the two sets have the samenumber of points. ◆

Note that this corollary implies that if F-1(q) is infinite for one point q inM, it is infinite for every point. Evidently, the exponential map E above hasinfinite multiplicity, as do the first two maps in Example 4.3.

A covering map of multiplicity 1 is just a diffeomorphism. A covering mapof multiplicity 2 is called a double covering. In Example 4.3 the last two mapsare double coverings. When faced with a choice of two objects at each pointof a surface M, it often turns out that the set of all these objects forms adouble covering surface of M.

a

M

MM


Corollary 4.5 can also be used to prove that a particular map is not a covering map. For example, let E1 be the restriction of the map E above toan interval J: 0 < t < 3p. Now J is still mapped nicely around the unit circleC - S1; in fact, E1: J Æ S1 is a local diffeomorphism. However E1 is not acovering map since for some points, F-1(q) contains two points, for othersonly one. A sketch will show that the even covering condition fails onlyaround the two edge points E1 (0) = (1, 0) and E1 (3p) = (-1, 0).

Proposition 4.4 has an analogue that asserts that 2-segments can also belifted to any level.

4.6 Theorem Let F: Æ M be a covering map, and let x: R Æ M be a2-segment, where D: a � u � b, c � v � d. If p is any point of such thatF(p) = x(a, c), then there is a unique 2-segment in such that

The proof is a straightforward 2-dimensional version of the proof of Propo-sition 4.4. The rectangle R is chopped into subrectangles, each lying in anevenly covered neighborhood. Then these are uniquely lifted—across one rowand back the next—until is completed. For details, see the Covering Homo-topy Theorem in [ST].

4.7 Remark Coverings with finite multiplicity. Suppose F: Æ M is acovering with multiplicity k.

(1) If M is compact, then is compact. Proof. M is covered by a finitenumber of 2-segments. Each of these can be lifted to k 2-segments in .Clearly these finitely many lifts cover all of , so it is compact.

(2) The Euler characteristic of is k times of that M. Proof. For a rec-tangular decomposition of D of M, lift (as in (1)) each face in D to k facesin . Using the uniqueness of curve lifts, we can check that if such facesmeet, they do so in a lifted edge or vertex. Each edge has k lifts, and the liftsof a vertex p are just the k points in F-1(p). Thus the lifted faces constitute arectangular decomposition of , and c( ) = kv - ke + kf = k(v - e +f ) = kc(M).

A covering surface can be expected to have simpler topology than thesurface it covers. Evidently the plane R2 is simpler than either a torus or acylinder. Now we show that simply connected covering surfaces are those thatcannot, in this sense, be further simplified.

MM

M

MM

MM

M

x

F a c˜ ˜ .x x x p( ) = ( ) =and ,

MxM

M



4.8 Theorem If F: Æ M is a covering map with connected and Msimply connected, then F is a diffeomorphism.

Proof. Since F is, in particular, a regular mapping onto M, it will sufficeto show that F is one-to-one. So if p1 and p2 are points of such thatF(p1) = F(p2) = q in M, we must show that p1 = p2.

Since is connected, there is a curve segment : [0, 1] Æ runningfrom p1 to p2. Now F carries to a curve a in M, and a is closed since

Because M is simply connected, a is homotopic to a constant. By Defini-tion 7.6 of Chapter 4, this means that there is a 2-segment x in M—definedfor simplicity on the unit square 0 � u, v � 1—whose base curve is a, withits other boundary curves constant at p.

By the preceding lemma there is a lift of x to the p1 level, that is, a 2-segment in such that

To show that p1 = p2, we chase around the rim of the 2-segment , asin Fig. 8.17.

The edge curve of starts at p1 and is carried by F to the constantcurve at p. Hence by the uniqueness of lifts (see proof of Prop. 4.4), canonly be the constant curve at p1.

Similarly the edge curve starts at (0) = (1) = p1 and is carried toa constant curve; hence is constant at p1.

Finally, ends at (1) = p1 and is carried to a constant curve; hence is also constant at p1. Thus

◆

Here are three applications of covering methods that involve orientationcovering surfaces (Exercise 6 of Section 4.8).

(1) Proof of Theorem 7.11 of Chapter 4. This asserts that a simply connected surface M is orientable. By the preceding theorem, the orientation

p p2 11 0= ( ) = ( ) =˜ ˜ .g b

bgbg

dgg

dxd

x

F ˜ ˜ .x x x p( ) = ( ) =and ,0 0 1

Mx

a a a a0 0 1 11 2( ) = ( )( ) = ( ) = = ( ) = ( )( ) = ( )F F F F˜ ˜ .p q p

aMaM

M

MM

FIG. 8.17

covering surface of M is not connected. For M Ã R3, it follows from Exer-cises 6 and 7 of Section 4.8 that M is orientable.

If M is not in R3 the argument is the same, but an abstract definition ofis required. For example, we could take to be the set of all rotation

operators J on M—two at each point.(2) The Poincaré-Hopf theorem for compact nonorientable surfaces N.

The proof in Section 7 of Chapter 7 required orientability; to extend to thenonorientable case, suppose that V is a vector field on N with only isolatedsingularities p1, . . . , pk.

Let F: Æ N be the orientation covering of N. Since F is a local diffeo-morphism, there is a unique vector field on N such that F*( ) = V.

For each pi, the two points qi, q¢i in F-1(pi) are also singular points of .Near each of them, is just an isometric copy of V near pi, so both qi andq¢i have the same index as pi. In view of the property of Euler characteristicin Remark 4.7,

Thus c(N) = Ski=1ind(V, pi).

(3) Classification of compact nonorientable surfaces. These surfaces canall be constructed by a scheme similar to that used in the orientable case(Section 7.6).

Instead of handles, crosscaps are used. A crosscap is a projective plane Pwith a hole punched in it. To add a crosscap to a surface M, punch a corre-sponding hole in it (in both cases, by removing the interior of a face of a rec-tangular decomposition). Join the boundaries of the holes smoothly to getthe new surface M¢. It has Euler characteristic

In fact, its decomposition has two fewer faces than M and P combined, andthe changes in vertices and edges cancel. (Recall that c(P) = c(S )/2 = 1.)

Then, analogous to Theorem 6.8 of Chapter 7, we have

4.9 Theorem If N is a compact connected nonorientable surface, thereis a unique integer k � 0 such that N is diffeomorphic to a projective planewith k crosscaps.

(See Chapter 1 of [Ma].) In this case we denote N by P[k]. The formula abovegives

c cP k P k k k[ ]( ) = ( ) - = - ( )1 0� .

c c¢( ) = ( ) -M M 1.

2 21 11

c cN N V V Vii

k

i ii

k

i

k

( ) = ( ) = ( ) + ¢( ) = ( )= ==

Â ÂÂˆ ˆ ˆ .ind , ind , ind ,q q p

VV

VVN

MM

M


Thus the nonorientable compact surfaces have Euler characteristics

The preceding theorem and Theorem 6.8 of Chapter 7 constitute the following surface classification theorem.

4.10 Theorem Two compact connected surfaces are diffeomorphic ifand only if they have the same Euler characteristic and are both orientableor both nonorientable. Equivalently, every compact connected surface is diffeomorphic to exactly one surface in the following double sequence:

The vertical arrows are orientation covering maps, since

4.11 Corollary The orientation covering surface of P[k] is S [k], a spherewith k handles.

Proof. If N is a nonorientable connected surface, then (as asserted inExercise 7 of Section 4.8) its orientation covering surface is connectedand orientable. If N is also compact, then by Remark 4.7, is compactand has Euler characteristic 2c(N). Thus the orientation covering ofP[k] is a compact orientable surface with Euler characteristic 2(1 - k) =2 - 2k. Hence by Theorem 6.8 of Chapter 7, P[k]

�is diffeomorphic to

S [k]. ◆

A local isometry of geometric surfaces that is also a covering map is calleda Riemannian covering map. To make use of the power of covering methods,it is important to be able to decide when a given local isometry is a Riemannian covering map. If it is, then curves in N can be lifted to every level in M. We now consider a geometrical converse.

A local isometry F: M Æ N is said to have the geodesic lift propertyprovided every geodesic segment g in N can be lifted to every level in M.(Note that such lifts are necessarily geodesics since the local inverses of F areisometries.)

4.12 Theorem If a local isometry F: M Æ N, with N connected, has thegeodesic lift property, then F is a Riemannian covering map.

NN

S S S S S= [ ] = [ ] [ ] [ ]Ø Ø Ø Ø= [ ] = [ ] [ ] [ ]>( ) =( ) <( )

0 1 2

0 1 2

0 0 0

, , , ,

, , , ,

T k

P P K P P P k

. . . . . .

. .. . . . .

c c c

k = - - -1 0 1 2 3, , , , , . . . .



Proof. To show that F maps M onto N, fix a point p0 in M and let q bean arbitrary point of N. Since N is connected, there is a broken geodesicsegment b: [0, b] Æ N from F(p0) to q (Exercise 7 of Section 7.4). Usingthe geodesic lift property, we lift successively the unbroken subsegments ofb to get a broken geodesic : [0, b] Æ M starting at p0. Thus

F( (b)) = b(b) = q.

We must find an evenly covered neighborhood for each point q in N. Itturns out that any normal e-neighborhood N of q will work. To show this,let p be any point of F -1(q). Each radial geodesic g starting at q runs forlength e, hence so does its lift to M starting at p. Since F is, in particu-lar, a local isometry, every geodesic starting at p is such a lift. Furthermore,once they leave p, no two such geodesic segments meet again. It followsthat these segments fill a normal e-neighborhood N (p) of p, and further-more that F carries N (p) in one-to-one fashion onto N. As a local isom-etry, F is a regular mapping; hence N (p) Æ N is a diffeomorphism.

Next we show that the neighborhoods N (p) for all p in F-1(q) fill theentire set F-1(N ) (Fig. 8.18). They are evidently contained in it. For r inF-1(N ), we must show that r is in some N (p). Let s be the radial geodesicfrom q to F(r) in N. Then if is the lift of s ending at r, we have

Hence (0) is in F-1(q), and r is in N ( (0)) as required.It remains only to note that for points p π p¢ of F-1(q), their normal

e-neighborhoods do not meet. In fact, if they did, then radial geodesic segments from p and p¢ meet—contrary to the uniqueness of lifts. ◆

ss

F ˜ .s s0 0( )( ) = ( ) = q

s

g

b

b

FIG. 8.18

Exercises

1. Describe a vector field on the projective plane that has exactly one sin-gularity. What is its index?

2. Let E: R Æ C be the exponential map in Example 4.2.(a) For any integer k, if Tk is the translation J Æ 2pk of R, show that ETk = E.(b) If N is the open semicircle y > 0 in the unit circle C, let cos-1 be the(differentiable) inverse of the restriction of cos to the interval I: 0 < J < pand define l: N Æ R by l(x, y) = cos-1(x). Prove that E(l(x, y)) = (x, y)for all (x, y) in N. (Hint: Use y > 0.)(c) Same as (b) for the semicircles y < 0, x > 0, and x < 0 in C.(d) Deduce that E is a covering map.

3. Let F: M Æ N be a local diffeomorphism. Show that F is a covering mapif F-1(q) contains the same finite number of points for all q in M.

4. For each integer n � 1, find a covering map Fn: C Æ C of the circle Cthat has multiplicity n.

5. Prove:(a) The only compact connected surface that can be a covering surface ofa torus T is a torus.(b) There are coverings F: T Æ T of every multiplicity n � 1.

6. Using sketches as needed, show:(a) A crosscap contains a Möbius band. (Hint: For F: S Æ P, remove thearctic and antarctic regions from S.)(b) If a crosscap is added to any surface, the resulting surface is nonorientable.(c) Cutting the Klein bottle in Fig. 8.20 by its plane of symmetry leavestwo Möbius bands.(d) P[1] is a Klein bottle. (Explain without using the classification theorems.)

8.5 Mappings That Preserve Inner Products

We have seen that a local isometry F: M Æ N carries geodesics of M to geodesics of N. The notation gv for the geodesic with initial velocity v allowsa more explicit description.

8.5 Mappings That Preserve Inner Products 425

5.1 Lemma If F: M Æ N is a local isometry, and v is a tangent vector toM, then

Proof. As just noted, = F(gv) is a geodesic of N. Its initial velocity isthe tangent vector

at the point F(p) in N. Thus by the uniqueness of geodesics, and gF*(v) arethe same. ◆

Thinking of an isometry as a rigid motion suggests that if two isometriesagree on some neighborhood, then they agree everywhere. In fact, a strongerresult is true.

5.2 Theorem Let F and G be local isometries from a connected surfaceM to a surface N. If at some one point p in M their differential maps agree,that is, if

then F = G.

Proof. If M is complete, the proof is easy. Let m be an arbitrary pointof M. By the Hopf-Rinow theorem (2.1) there is a vector v at the specialpoint p such that gv(r) = m for some number r. Then the preceding lemmagives

Hence, in particular,

In the general case, there is at least a broken geodesic b from p to m(Exercise 7 of Section 7.4). Then F(b) = G(b) follows by applying the argu-ment above, successively, to each unbroken segment of b. So again F(m) =G(m). ◆

An isometry F: M Æ M from a surface to itself can be regarded as anintrinsic symmetry of M. Every feature of the geometry of M is the same atp as at F(p), since this geometry consists of isometric invariants. The resultsof Exercise 9 of Section 6.4, show at once that the set I(M) of all isometries

F F r G r Gv vm m( ) = ( )( ) = ( )( ) = ( )g g .

F Gv F v G v vg g g g( ) = = = ( )( ) ( )* *.

F G T M T M F Gp q* *:= ( ) Æ ( ) = ( ) = ( ), where ,q p p

g

¢( ) = ¢( )( ) = ( )g g0 0F Fv* * v

g

F v F vg g( ) = ( )*.


M Æ M forms a group, just as does the set E (n) of all isometries of n-dimensional Euclidean space Rn (Exercise 7 of Section 3.1). I(M) is called the isometry group of M.

The group I(M) is intrinsic to M and when M is a surface in R3 should notbe confused with the group S(M) of Euclidean symmetries of M (Exercise 6of Section 6.9). Intuitively, S(M) gives the symmetries of M seen by Euclid-ean observers, while I(M) gives those observed by the inhabitants of M. EachEuclidean symmetry restricts to an isometry F |M : M Æ M, but this does notgenerally give all the isometries of M Ã R3, as we now see.

5.3 Example Let M be a cylinder in R3 whose cross-sectional curve is anoncircular ellipse of length l. Parametrize M by

where a is a periodic unit-speed parametrization of the ellipse. Now for anynumber a, the map

is an isometry of M. The inhabitants of M cannot distinguish M from theisometric cylinder C whose cross-sectional curve is a circle of circumferencel; so for them, Fa is just a rotation.

But evidently Fa will be the restriction of a Euclidean symmetry F only inthe special case where a is a multiple of l/2. Then F is a 180° rotation of Mor the identity map.

For an arbitrary geometric surface M, the isometry group I(M) gives anovel algebraic description of M. Roughly speaking, the more symmetricalM is, the larger I(M) is. For example, the ellipsoid

has eight isometries derived from Euclidean isometries: three reflections (one in each coordinate plane), three 180° rotations (one around each coordinate axis), the isometry p Æ -p, and, of course, the identity map ofM. In fact, these are the only isometries of M (argue from K as in Exercise8 of Section 6.9).

The smallest possible isometry group I(M) occurs when the identity map is the only isometry of M. We can produce such a surface by putting abump on the ellipsoid in such a way as to destroy all seven of its nontrivialisometries.

Mxa

yb

zc

a b c:2

2

2

2

2

21+ + = > >( )

F u v u a va x x, ,( )( ) = +( )

x u v u vU, ,( ) = ( ) +a 3


By contrast, a surface M has the maximum possible symmetry if everyisometry allowed by Theorem 5.2 actually exists. Explicitly, given any linear isometry f : Tp(M) Æ Tq(M) of tangent spaces, there exists an isometry F: M Æ M such that F* = f.

Equivalently, given frames e1, e2 and 1, 2 at p and q, respectively, thereexists an isometry F: M Æ M such that

Thus in this case we say that M is frame-homogeneous: Any two frames onM are symmetrically positioned. So what was proved in Theorem 2.3 ofChapter 3 is that R3 is frame-homogeneous, and the same proof is valid forany Rn, in particular, for the Euclidean plane R2.

5.4 Definition A geometric surface M is point-homogeneous (or merelyhomogeneous) provided that for any points p and q of M there is an isome-try F: M Æ M such that F(p) = q.

A frame-homogeneous surface is, of course, homogeneous—but not con-versely. A circular cylinder C furnishes an example. Rotations of R3 aboutthe axis of C and translations of R3 along this axis produce isometries of C.Evidently we can move any point of C to any other using a rotation and atranslation, so C is homogeneous. But C is not frame-homogeneous: All itspoints are geometrically equivalent, but not all its frames. Proof. No isome-try F could carry any vector v tangent to a ruling to a vector w tangent to a cross-sectional circle, since by Lemma 7.1, F would have to send the one-to-one geodesic gv to the closed geodesic gw an impossibility since F isone-to-one.

Homogeneity is a strong restriction.

5.5 Theorem If a geometric surface M is homogeneous, then M is com-plete and has constant Gaussian curvature.

Proof. Constancy of curvature follows immediately from the definitionof homogeneity and the fact that isometries preserve curvature.

The proof of completeness is more interesting. If M is not complete,there is a unit-speed geodesic a whose largest interval of definition I is notall of R. Suppose I: t < b. Let us show that this is impossible.

The radial geodesics in a normal e-neighborhood of some point p allrun for length e > 0. Choose t0 in I so that b - t0 < e. Because M is homo-

F F* * .e e e e1 1 2 2( ) = ( ) =,

ee



geneous, there is an isometry F: M Æ M such that F(p) = a(t0). For someunit vector u at p, F*(u) = a ¢(t0). Thus the geodesic segment F(gu) has initialvelocity

and this segment runs for distance e at unit speed (Fig. 8.19). By theuniqueness of geodesics, a reparametrization of gu can be attached onto ato produce an unbroken geodesic defined on the interval I¢: t < t0 + e. Sincet + e > b, this contradicts the maximality of the interval I and thus provesthat M is complete. ◆

As the title of this section suggests, isometries and local isometries are notthe only inner-product preserving mappings of importance in geometry. Weconsider briefly some other types.

5.6 Definition Let F: M Æ Rn be a mapping of a geometric surface intoEuclidean n-space (n = 3, 4, . . .). If the differential map F* preserves innerproducts of tangent vectors, then F is an isometric immersion.

If, furthermore, F is one-to-one and its inverse F-1: F(M) Æ M is continu-ous, then F is an isometric imbedding.

For example, a proper patch (Section 1 of Chapter 4) is an imbedding ofan open set D Ã R2 into R3. The continuity requirement for the inverse mapF-1 is the same as in that special case: It prevents edges of M from being gluedonto M, thus making F(M) quite different from M. In the special case wherethe surface M is compact, it has no “edges,” and in fact a theorem of topol-ogy asserts that F-1 is always continuous.

The following technical result makes it clear that the surfaces M Ã R3

studied in the early chapters are just those abstract geometric surfaces thatcan be imbedded in R3.

F F tu* *¢ ( )( ) = ( ) = ¢( )g a0 0u ,

FIG. 8.19

5.7 Lemma If F: M Æ R3 is an isometric imbedding of a geometricsurface in R3, then the image F(M) is a surface in R3 (Definition 1.2 ofChapter 4), and the function F: M Æ F(M) is an isometry.

Proof. If x: D Æ M is a coordinate patch in M, then the compositemapping F(x): D Æ R3 is a proper patch. In fact, its inverse functionF(x(D)) Æ D is x-1F-1, which is continuous since both x-1 and F-1 are.Evidently, F(x(D)) is contained in M, so the definition of surface in R3 issatisfied.

F(M) uses the dot product of R3 as its inner product, and by definitionthe imbedding F: M Æ R3 preserves inner products. Hence, when consid-ered as a mapping of M onto F(M), F also preserves inner products, andsince it is one-to-one, it is an isometry. ◆

We have seen that there are geometric surfaces M that cannot be imbeddedin R3, for example, the flat torus and the projective plane (both defined inSection 2 of Chapter 7). In such cases it is natural to try to imbed M in ahigher-dimensional Euclidean space Rn. Finding isometric imbeddings isseldom easy, but the larger n is, the less difficult the task becomes. Roughlyspeaking, with more dimensions for M to curve through, there is a betterchance that a shape can be found for it that is compatible with its intrinsicgeometry.

The following example imbeds the flat torus; for the projective plane seeExercise 10.

5.8 Example Isometric embedding of a flat torus in R4. The idea here isthat since the unit circle C is naturally imbedded in the plane R2, by takingCartesian products (Exercise 15 of Section 4.8) we can get an imbedding ofthe flat torus C ¥ C = T0 into R2 ¥ R2 = R4.

Start with the mapping : R2 Æ R4 given by

If x is the parametrization of the flat torus given in Example 2.2 of Chapter7, then the formula

is consistent in the sense of Exercise 13 of Section 4.5. Now

x x

x x

u v u v

u m v v n

u v u v

, ,

u ,

, , ,

( ) = ( )¤ = + = +

¤ ( ) = ( )

1 1

1 1

1 1

2 2p p

F u v u vx x, ,( )( ) = ( )

x u v u u v v, , ,( ) = ( )cos sin cos , sin .

x


Thus the exercise shows not only that F is a well-defined mapping F: M Æ R4, but also that it is one-to-one.

To show that F preserves inner products, the method in Lemma 4.5 ofChapter 6 remains valid. We compute

Hence, using the dot product of R4,

Since these functions agree with E, F, G for x, inner products are preserved. ◆

The general situation here is not well understood. It is known that everycompact surface can be isometrically imbedded in R17, but it seems likely that17 can be replaced by a lower dimension.

Exercises

1. Prove that isometric surfaces have isomorphic isometry groups.

2. Prove:(a) The torus of revolution T is not homogeneous.(b) The flat torus T0 is homogeneous, but not frame homogeneous. (Hint:Show that T0 has closed geodesics of different lengths.)

3. Suppose that in M any two points can be joined by at least one geodesic,and that in N any two points can be joined by at most one geodesic. Provethat every local isometry F: M Æ N of such surfaces is one-to-one.

4. Show that the set I + of all orientation-preserving isometries of a surfaceM is a normal subgroup of the isometry group I(M). (A subgroup H of agroup G is normal provided h Œ H and g Œ G imply ghg-1 Œ H.)

5. (a) If there is a point p0 in M such that for each p in M some isometryof M sends p0 to p (or the reverse), show that M is homogeneous.(b) State and prove the analogous fact for frame-homogeneity.(c) Suppose there is a point p0 in a homogeneous surface M with the prop-erty that for any two frames at p0 there is an isometry F of M such that F*carries one of the frames to the other. Show that M is frame-homogeneous.

E F G= = =1 0 1, , .

x

x

u

v

u u

v v

= -( )= -( )

sin cos

sin cos .

, , , ,

, , ,

0 0

0 0


6. For the sphere S: ||p|| = r in R3, prove:(a) Every orthogonal transformation C of R3 restricts to an isometry of S.(b) S is frame-homogeneous.(c) Every isometry of the sphere is the restriction of a Euclidean isome-try. (Hint: For (b) use Ex. 5(c).)

7. Prove:(a) If C: S Æ S is an isometry as in Exercise 6(a), there is a unique isom-etry Cp: P Æ P such that FC = CpF, where F is the projection S Æ P.(b) The projective plane P is frame-homogeneous.

8. If M is a connected surface in R3 that is not contained in a plane, showthat the function F Æ F|M is a one-to-one homomorphism of the Euclideansymmetry group S(M) onto a subgroup of the isometry group I(M).

9. Let M be the trough-shaped surface z = cosh2y in R3.(a) By adjusting the Monge patch (u, v) Æ (u, v, coshv) find an isometryx: R2 Æ M.(b) Show that I(M ) is isomorphic to the isometry group E (2) of the Euclid-ean plane. Which isometries of M derive as in Exercise 8 from Euclideansymmetries of M? Which do not?

10. (Isometric imbedding of the projective plane.) Consider the mapping F: R3 Æ R6 given by

(a) If v is a tangent vector to R3 at p = (x, y, z), show that

Thus the restriction F1 = F |S of F to the unit sphere S is an isometricimmersion of S in R6.(b) Show that F1 is one-to-one, hence is an imbedding. (By a theorem of topology, since S is compact, the inverse map (F |S)-1 is automaticallycontinuous.)(c) Derive an isometric imbedding of the projective plane P in R6. (Hint:Prop. 2.6 of Ch. 7.)(d) Check that each point in the image of F1 has the same dot product with (1, 1, 1, 0, 0, 0), and improve (c) to an isometric imbedding of P in R5.

11. The n-dimensional sphere S n: ||p|| = r in Rn+1 becomes a Riemannianmanifold, just as in the 2-dimensional case, by using the dot product of Rn+1

on its tangent vectors. Show that:

F* • .v p v p v( ) = + ( )2 2 2 2

F x y z x y z xy xz yz, ,( ) = ÊË

ˆ¯

1

2

1

2

1

22 2 2, , , , , .


(a) The mapping in Example 5.8 actually gives an isometric imbedding ofthe flat torus in a 3-dimensional sphere of radius .(b) The mapping in the preceding exercise gives an isometric imbedding ofthe projective plane in a 5-dimensional sphere of radius .

8.6 Surfaces of Constant Curvature

The simplest possibility for the Gaussian curvature of a surface is that it isconstant. We have seen several examples of such surfaces, and there are infi-nitely many more. The goal of this section is to give a reasonable organiza-tion for all of them.

A realistic treatment must be limited to complete connected surfaces.Without connectedness, we would have to consider, for example, a surfacecomposed of a random collection of planes, spheres, and tori. Without com-pleteness we would have to deal with enormous collections starting, say, withevery connected open subset of the plane.

Given any number k, there is a particularly simple geometric surface M(k)whose Gaussian curvature has the constant value k.

• k > 0: M(k) is the sphere S Ã R3 of curvature k (hence radius ).• k = 0: M(k) is the Euclidean plane R2.• k < 0: M(k) is the hyperbolic plane H of curvature k (hence pseudo-radius

). (See Exercise 4 of Section 7.2.)

We call M(k) the standard geometric surface of constant curvature k. Thesesurfaces are complete and simply connected, and using covering methods wewill show that in a sense they dominate all surfaces of constant curvature.

The following preliminary result will simplify matters.

6.1 Lemma Let M be a complete connected surface and let F: M Æ Nbe a local isometry into a connected surface N. Then F is a Riemannian cov-ering map onto N, and N is complete.

Proof. If the geodesic lift property holds, then Theorem 4.12 will showthat F is a Riemannian covering map. So let a: [0, b] Æ N be a geodesicsegment and let p be a point of M such that F(p) = a(0). There is a uniquevector v at p such that F*(v) = a ¢(0). Since M is complete, the geodesic gv

can be defined on the entire real line. Then

F v F vg g a( ) = =( )*,

r k= -1/

1/ k

1 2/

2

8.6 Surfaces of Constant Curvature 433

wherever both sides are defined. Thus gv provides a lift of a, and furthermore, F(gv) is an extension of a over the whole real line, so N iscomplete. ◆

6.2 Theorem If N is a complete connected surface with constant curva-ture k, then the standard surface M(k) is a Riemannian covering surface ofN.

Proof. In view of the preceding lemma we need only prove that there isa local isometry F from M(k) into N. There are three cases.

The Case k < 0. It will suffice to work with the k = -1 hyperbolic planeH.

As in (2) of Example 1.7, we use the frame U1, U2 at 0 in H. Let e1, e2

be a frame at an arbitrary point of N. Then let and x be the resultinggeodesic polar mappings in M(k) and N. For the surface N we assert:

(1) x is defined on the entire closed half-plane H : u � 0 (a consequenceof completeness).

(2) The mapping x: H Æ N is regular for u > 0. (Proof: As noted in theproof of Lemma 1.6, geodesic polar mappings have E = 1 and F = 0.We saw earlier that the Jacobi equation for k = -1 gives ,so EG - F 2 = sinh2 u > 0 for u > 0.)

These general results are valid for : H Æ H as well, but here we knowmore. Example 1.7 showed that the whole surface H is a normal neigh-borhood of the pole 0. Thus has only the usual ambiguities of polarcoordinates, that is, the equation (u, v) = q determines u and v uniquely,but for the addition of multiples of 2p. This means that the formula

is consistent in the sense of Exercise 13 of Section 4.5. Thus it defines amapping F of H onto N. Condition (1) ensures that F is defined on all ofM(k), and the differentiability of F at the pole follows from the differen-tiability of the exponential map there.

That F is a local isometry follows from the fundamental criterion inSection 4 of Chapter 6. Indeed, using (2) above gives

At the pole the preservation of inner products is an honest consequenceof continuity.

The Case k = 0. This proof is a word-for-word copy of the precedingone, except that M(k) = R2 and = G = u2.G

˜ ˜ ˜ sinh .E E F F G u G u= = = = = = >1 0 02, , for

F u v u vx x, ,( )( ) = ( )

xx

x

G u= sinh

x


The Case k > 0. Here a new idea is required, since the largest normalneighborhood N of a point p in the sphere M(k) = S is not the wholesphere, but the sphere minus the antipodal point -p of the pole p.

Arguing as in the case k < 0, we get a local isometry F1: N Æ N. Nowrepeat this argument once more at a point p* different from both p and -p. This will produce another local isometry F2: N * Æ N, where N * isall of S except -p*. For the frames that determine F2 we use an arbitraryframe e1, e2 at p*, but in N the frame F1*(e1), F1*(e2) at F1(p*).

Thus Theorem 5.2 applies, showing that F1 = F2 on the (connected) inter-section of N and N *. But N and N * cover the entire sphere, so takentogether F1 and F2 constitute a single local isometry F from S to N. ◆

6.3 Corollary The standard constant curvature surfaces M(k) are theonly geometric surfaces that are complete, simply connected, and have con-stant curvature.

Proof. We have already seen that the three types of M(k) have the spec-ified properties. Conversely, if M is a complete connected surface with con-stant curvature k, then by Theorem 6.2 there is a Riemannian covering mapF: M(k) Æ M. But Theorem 4.8 asserts that a simply connected surfacehas only trivial coverings, so F is an isometry. ◆

Using Theorem 6.2 we give an overview of constant curvature surfaces,omitting some proofs. We assume connectedness for all surfaces, and con-sider the cases k > 0, k = 0, k < 0 in turn.

Constant Positive Curvature Given any number k > 0 there are (up toisometry) exactly two complete geometric surfaces M with curvature k: thesphere and the projective plane. The preceding corollary covers the simplyconnected case. Since the sphere has Euler characteristic 2, Remark 4.7implies that it can nontrivially cover only a compact surface of Euler char-acteristic 1. By Theorem 4.10, the projective plane P is the only such surface.

Flat Surfaces For k = 0 we have met, so far, three types of complete sur-faces: the plane, cylinder, and flat torus. There are two more types, discov-ered only in the late 1800s. They can be found by the scheme used to derivethe projective plane P from the sphere S. Example 8.2 of Chapter 4 used theantipodal map A of the sphere to construct P as an abstract surface; thenExample 2.7 of Chapter 7 carried geometry along to make P a geometricsurface. Starting from a surface M (in the role of S), all that is needed is anisometry A of M that has the two essential properties of the antipodal mapof S:



A2 = I (the identity map of M), andA has no fixed points, that is, A(p) never equals p.

Then the abstract surface M/A is defined to be the set of all unorderedpairs {p, A(p)}, and patches are defined just as for the projective plane P = S /A. Evidently, Proposition 2.6 of Chapter 7 applies, giving M/A ametric tensor that makes the projection p Æ {p, A(p)} a local isometry. Then,by Lemma 6.1, it is a Riemannian covering.

6.4 Example Complete flat nonorientable surfaces.(1) A complete flat Möbius band. Let C be the cylinder x2 + y2 = r2 in R3.

For the required isometry A we use the antipodal map A(x, y, z) = (-x, -y,-z). It may help to recognize the resulting surface C/A as a Möbius band byimagining the cylinder as shrunk down to an ordinary band, its vertical linesreduced to intervals -1 < z < 1.

However, we need the whole cylinder C, so that it, and hence C/A, will becomplete.

(2) A flat Klein bottle. The familiar image of the Klein bottle as labora-tory glassware (right side of Fig. 8.20) is not very flat. But we can constructa flat one, starting from the flat torus T0 in Example 2.2 of Chapter 7.

T0 has the same symmetries in R3 as the torus of revolution, so once more,the mapping A(x, y, z) = (-x, -y, -z) is an isometry T0 Æ T0. Then T0/A isa complete flat surface.

The transition in Fig. 8.20 suggests how to see that T0/A is actually a Kleinbottle. On the left we have discarded an open half of T0 since the deletedpoints are antipodal to points in the remaining tube (hence nothing is lost inthe construction of T0/A). The identifications imposed by A on the two

FIG. 8.20

boundary circles give them orientations inconsistent from those that wouldproduce a torus. To match these oriented circles, the tube must be stretchedaround as suggested in Fig. 8.20 and twisted along the way. The criterion inExercise 1 of Section 4.8 shows readily that T0/A is nonorientable.

It is known that there are no more. Of course, except for the plane, differentparameters produce nonisometric examples within a given type, for example,the cylinders over circles of different radii.

All five types appear in the following diagram, where arrows represent Riemannian covering maps.

For the horizontal arrows, the multiplicity of the covering is infinite. The ver-tical arrows are the two double coverings defined above.

Only the flat torus and Klein bottle are compact; only the Möbius bandand Klein bottle are nonorientable. In fact, both vertical arrows represent ori-entation coverings.

Only the plane and cylinder can be isometrically imbedded in R3. Compactflat surfaces are barred from R3 by Theorem 3.5 of Chapter 6. For the (non-compact) flat Möbius band, see the reference in Exercise 4 of Section 2.

Constant Negative Curvature Ignoring scale changes, we have foundvery few surfaces that have constant k � 0, but the situation is radically dif-ferent for k < 0.

6.5 Theorem Every compact surface with a negative Euler characteris-tic admits a metric of constant negative curvature.

(Of the infinitely many compact surfaces listed in the classification theorem(4.10), all but four have c < 0.)

We give an informal proof of the theorem in the orientable case. Considera regular geodesic octagon P in the k = -1 hyperbolic plane, as shown in Fig.8.21. P is centered at 0, and regular means that its eight angles are equal andalso its eight sides. Recall that the flat torus defined in Example 2.2 ofChapter 7 was visualized as a rectangle with opposite sides sewn together.

plane cylinder flat torus

Mobius band Klein bottle

Æ ÆØ Ø

˙

So now up to scale change we have five types of complete flat surfaces

Euclidean plane, cylinder, flat torus, Mobius band, Klein bottle

— — .

˙ .



Figure 8.21 uses the same scheme: Like-named sides are sewn together withmatching orientations (indicated by the arrows). When this is done, the eightvertices of P are fused into one. All the angles of P meet there; hence if theresult is to be a surface, their sum must be 2p.

This can always be achieved by adjusting the size of P. To see this, let Pr

be the regular octagon that has hyperbolic distance r from 0 to every vertex.The Gauss-Bonnet formula proves what can be seen in figures such as Fig.7.11: for r small, the angle sum is too large, and for r very large, the sum istoo small. (Recall that hyperbolic angles are the same as Euclidean.) Sincethe angle sum depends continuously on r, there is an intermediate value forwhich the sum is exactly 2p.

Let be the resulting surface. It is compact and since it is cut from thehyperbolic plane has constant negative curvature. We claim that topologicallyit is a double torus, that is, a sphere with two handles. In Fig. 8.21, note thatthe line from vertex to vertex becomes a circle in . On the right side the fouredges a, b, a, b are sewn together by the pattern used for the torus. Thus theright half of the figure produces in a torus with a hole punched in it. Thesame is true for the left side, so is constructed by joining two handles; it isthus a double torus, that is, a sphere with two handles.

A proof by induction takes care of the general case. To reach the spherewith h > 2 handles, start with a regular 4h-sided polygon and modify Fig.8.21 by replacing sides c, d, c, d (on the left) by identification pattern of thesphere with h - 1 handles. Then, for sides a, b, a, b (on the right) add onemore handle just as before. ◆

It turns out that two compact surfaces with the same curvature k < 0 andthe same number h � 2 of handles need not be isometric. However, they musthave the same area A(k, h) = 4p(1 - h)/k, since by the Gauss-Bonnettheorem,

PP

P

P

FIG. 8.21


If compactness is weakened to completeness, there are infinitely many non-isometric constant k < 0 surfaces, with complicated patterns of tubes andspikes that defy any systematic organization.

As a simple example, let M be the abstract surface built by attaching aninfinite number of handles to a cylinder, as in Fig. 8.22. Let F be a coveringmap of M onto a double torus D that wraps the cylinder repeatedly aroundthe lower loop L of D in such a way that each handle of M is carried dif-feomorphically onto the upper loop U of D. We have just seen that D has ametric with constant negative curvature. The pullback metric on M makes F a local isometry (Remark 1.3 of Section 7) and thus gives M constant negative curvature.

Not one of these many k < 0 surfaces can be found in R3, for a theoremof Hilbert asserts that a complete surface in R3 cannot have constant nega-tive curvature; this is proved in [dC]. A refinement by Efimov weakens thecurvature hypothesis to K � k < 0.†

Scale changes effectively reduce the standard surfaces M(k) to just three:the unit sphere S , Euclidean plane R2, and hyperbolic plane H of curvaturek = -1. The plane, in concept at least, was well known long before Euclid.The sphere was prominent in ancient astronomy and later in long-distancenavigation. Only hyperbolic geometry was a self-conscious mathematicalinvention.

Like the Euclidean plane, the hyperbolic plane was initially studied syn-thetically, that is, in terms of axioms governing relations between abstractpoints and lines. Only later were these synthetic geometries realized by geo-metric surfaces whose geodesics are the “lines.” The familiar axiom that twopoints determine a unique line fails for antipodal points on the sphere S , but

kA K dM M h h= = ( ) = -( ) = -( )ÚÚ 2 2 2 2 4 1pc p p .

FIG. 8.22

† See T. K. Milnor, “Efimov’s Theorem about Complete Immersed Surfaces of Negative Curvature,”Advances in Mathematics 8 (1972), 472–543.

it is recovered by identifying antipodal pairs, thus forming the projectiveplane P.

The classical geometries are those of the Euclidean plane, sphere, projec-tive plane, and hyperbolic plane. In the sphere any two lines (that is, routesof geodesics) must intersect; hence the same is true for P. Thus the parallelpostulate for these four geometries becomes:

Through any point p not on a line L, the number of lines through p thatdo not meet L is

6.6 Corollary All four surfaces R2, S, P, H realizing the classical geome-tries are frame-homogeneous.

Proof. We already know this for M = R2, S, and H, the latter two fromearlier exercises. But Theorem 6.2 gives a simple proof for all three simul-taneously. Fix a tangent frame to M—for H locate it at the origin 0. Bythe proof of Theorem 6.2 this frame can be carried to an arbitrary frameon M by a Riemannian covering map. The surfaces M are simply con-nected; hence Theorem 4.8 implies that this covering map is an isometry.This is enough to prove frame-homogeneity (see Exercise 5 of Section 5).

Since the projective plane is not simply connected, it escapes the argu-ment above. But its frame-homogeneity can be derived from that of thesphere, as suggested in Exercise 7 of Section 5. ◆

Previously, as in this proof, we have given preference to the origin 0 in thehyperbolic plane H, but now we know that all frames on H are geometricallyequivalent.

Frame homogeneity is sometimes called the axiom of free mobility becausethe inhabitants of such a surface can then use (as we have) Euclid’s intuitivedefinition of congruence: Two geometric figures are congruent provided onecan be moved isometrically into coincidence with the other.

The surfaces in the preceding corollary are the only frame-homogeneoussurfaces. In fact, if k > 0, then M is either the sphere or projective plane, bothframe-homogeneous. If k = 0, then among the five flat surfaces only the planeis frame-homogeneous (see Exercise 2 of Section 5 and Exercise 2 below).More advanced covering methods are needed in the k < 0 case; there thehyperbolic plane is the only surface that is even homogeneous.

It is noteworthy that of the multitude of complete, constant curvature surfaces only three types—spheres, cylinders, and the plane—can be foundin R3.

hyperbolic , Euclidean: , spherical and elliptic: : .• 1 0


Exercises

1. (a) How do we know, before trying it, that the construction method forTheorem 6.5 cannot produce a S [1] (sphere with one handle) that has con-stant k = -1?

(b) If we do try this method, exactly where does it fail?

2. Show that the flat Möbius band M contains a unique shortest closed geo-desic. Deduce that M is not homogeneous (hence not frame-homogeneous).

3. An incidence geometry consists of two abstract sets—called the pointsand the lines—and a single incidence relation described by either “point is online” or “line is through point.” Consider the following axioms:

(i) Through any two distinct points there is a unique line.(ii) Any two distinct lines pass through a unique point.(iii) There exists a set of four points, no three of which are on the sameline.

For each of the classical geometries M = R2, S, P, H decide which of theseaxioms are satisfied, with points of M as the points, routes of geodesics as lines,and the obvious incidence relation. Which (if any) satisfy all three axioms?

4. In the sphere S of radius r, let D be a triangle whose sides are geodesicsegments of lengths a, b, and c (all less than pr). Let J be the interior angleof D at the vertex p opposite side a.

(a) Prove this spherical law of cosines:

(Hint: To determine cosJ, find unit vectors at p tangent to the sides b andc.)(b) Show that this formula approximates the usual Euclidean law ofcosines when r is large compared to a, b, c.

5. (Side-angle-side criterion). Let D and D¢ be geodesic triangles in the hyper-bolic plane H. If two adjacent sides a, b of D have the same lengths as cor-responding sides in D¢ and the angles between the sides are equal, prove thatthe triangles are congruent, that is, show that there exists an isometry F ofH such that F(D) = D¢.

6. Show: (a) Every S [n], n � 2, is a covering surface of the double torusS [2].

(b) S [4] can be covered by S [n] if and only if n has the form 3N + 1. (Hint:To show existence, adapt Fig. 8.22.)

cos cos cos sin sin cos .ar

br

cr

br

cr

= + J


8.7 Theorems of Bonnet and Hadamard

The simplest way to weaken the hypothesis that Gaussian curvature is con-stant is to require that it obey inequalities. The natural cases may seem to beK < 0 and K > 0, but, in fact, they turn out to be K � 0 and K � k > 0. Thereason stems from the opposite effects produced by the Jacobi equation: Aswe have seen, K � 0 implies that there are no conjugate points, and, as wenow show, K � k > 0 implies that there are many.

The following is an instance of the general rule that conjugate points arrivesooner on surfaces with larger positive curvature.

7.1 Lemma Let M be a surface with K � k > 0 for some constant k. Ifa geodesic segment s starting at p has length , there is a conjugatepoint of p along s.

Proof. The sphere S of radius has curvature k = 1/r2, so Mhas curvature at least as large as that of S . We know that in S , conjugatepoints arrive at arc length , so the lemma asserts that they arriveno later on M.

To prove this, let s be a unit-speed geodesic in M defined on the inter-val [0, l], where . From Corollary 3.4, the Jacobi equation andinitial conditions for M are

and the corollary asserts that there will be a conjugate point of s(0) alongs if and only if the function g is zero at some s1 with 0 < s1 £ l.

The corresponding data for S are

Here we have the explicit solution

We will show that as long as g(s) stays positive, it is no larger than f(s).Since f is zero at l, it follows that g has a zero at or before l, thus provingthe lemma.

The assertion is proved by some simple but ingenious calculus: Supposethat g > 0 on the open interval J: 0 < s < b, where b � l. Evidently f > 0on J, so the Jacobi equations above give

f sks

k( ) =

sin.

¢¢ + = ( ) = ¢( ) =f kf f f0 0 0 0 1, with , .

¢¢ + ( ) = ( ) = ¢( ) =g K g g gs 0 0 0 0 1, with , ,

l p= � K

p pr K= �

r K= 1�

� �p K


8.7 Theorems of Bonnet and Hadamard 443

Thus the function gf ¢ - fg¢ is nondecreasing (that is, monotonically increas-ing) on J. Since gf ¢ - fg¢ is zero at s = 0, it follows that gf ¢ - fg¢ � 0 onJ. Hence

So f/g is also nondecreasing on J. This function is equal to 1 at s = 0, sinceby 1’Hôpital’s rule,

Thus f/g ≥ 1 on J, that is, f � g there. ◆

The main theorem on curvature K � k > 0 continues the comparison witha sphere S of curvature k and shows that M also is compact—and has diam-eter and area no larger than those of S. “Diameter” here means intrinsicdiameter: the least upper bound of all distances r(p, q) between points ofM—or • if these distances are unbounded. For a surface in R3, this shouldnot be confused with the Euclidean diameter, which is generally smaller, sinceEuclidean distances tend to be shorter (see Fig. 6.8).

For example, consider the sphere S of curvature k. It has Euclidean radius, hence intrinsic diameter , the distance between antipo-

dal points. Its Euclidean diameter, of course, is just . Note that interms of curvature its area 4pr2 becomes 4p/k.

7.2 Theorem (Bonnet) If M is a complete connected surface with K � k > 0, then M is compact and has intrinsic diameter and area � 4p /k.

Proof. If we show that for all p, q in M, then the diam-eter inequality follows.

By the Hopf-Rinow theorem there is a minimizing geodesic segment sfrom p to q. Then s is certainly locally minimizing, so Theorem 3.7 assertsthat there are no conjugate points of p on s before q.

By the preceding lemma, there will be a conjugate point of p on sif s is strictly longer than . Thus . Since s is minimizing,

as required.

r sp

p q,( ) = ( )Lk

� ,

L ks p( ) � /l p= / k

r pp q,( ) � / k

� p / k

2 2r k= /p pr k= /r k= 1 /

lim lim .s s

f sg s

f sg sÆ Æ

( )( ) =

¢( )¢( ) =

0 01

fg

g f fgg

ÊËÁ

ˆ¯

¢=

¢ - ¢2

0�

gf fg gf fg fg K k J¢ - ¢( )¢ = ¢¢ - ¢¢ = -( ) � 0 on .

To show that M is compact, consider the exponential map exp at anarbitrary point p. Line segments radiating from 0 in the tangent planeTp(M) are carried by exp to radial geodesics—of the same length—startingat p. The argument above shows that we can travel from p to any point qof M by a geodesic of length at most . Thus exp maps the diskD: ||v|| � l in Tp(M) onto the entire surface M. The disk D is compact,and the continuous image of a compact set is compact (Exercise 2 ofSection 4.7); hence M is compact.

For the area assertion, we only need to integrate over enoughof the geodesic polar parametrization at p to be sure of covering all of M.(No harm will result if some of M is covered more than once.) Let s1(v)be the distance to the first conjugate point of p along the radial geodesics Æ x(s, v), where we write s instead of the usual u. The proof of thepreceding lemma showed that on the interval Jv: 0 < s < s1(v),

is never greater than

Since E = 1 and F = 0 for a polar parametrization,

on the interval Jv.

The argument above shows that to cover all of M it suffices to go outto s1(v) on each radial geodesic. Then, since remains positive outto ,

◆

We saw in Section 2 that compactness implies completeness. Bonnet’stheorem provides a converse when K � k > 0. Just K > 0 would not suffice—for example, the paraboloid of revolution is complete and has K > 0, but iscertainly not compact.

Bonnet’s theorem can be strengthened as follows.

area M dvs

k

kdv

s

kds

k

kk

ksk

k

ds

s ds

s v k

kk

� �0

2

0

1

00

2

00

2 2

4

p pp

pp p

p

p

Ú Ú

Ú

( ) ( )

= ( ) = - ( )[ ]

=

( )

ÚÚsin sin

cos

.

/

/sin /

p / k s v� 1( )sin sk( )

EG F s v G s v g sk

ksv- ( ) = ( ) = ( ) ( )2 , , �

sin

sin .

( )k

ks

g s G s vv( ) = ( ),

EG F- 2

l p= / k


7.3 Corollary Under the hypotheses of Bonnet’s theorem, M is diffeo-morphic to either a sphere or a projective plane, and in the latter case thebounds on diameter and area can be cut in half.

Proof. M is compact, connected, and has positive curvature; hence, it haspositive Euler characteristic. Thus the first assertion is an immediate con-sequence of the surface classification theorem (4.10). When M is dif-feomorphic to a projective plane, its orientation covering surface is diffeomorphic to a sphere. If F: Æ M is the covering map, assign the pullback metric, thus making F a local isometry. Now apply Bonnet’stheorem to . Since this is a double covering, it is clear that the area ofM is exactly half that of . We omit the more complicated proof of thecorresponding fact for diameters. ◆

Turning to the study of a surface M with K � 0 we again use compari-son—this time with the Euclidean plane in the form of any tangent plane toM. The essential fact is that the exponential maps of M are covering maps.To prove this, two lemmas are needed.

7.4 Lemma If M is a complete surface with K � 0, then for any point pin M the exponential map expp: Tp(M) Æ M is length nondecreasing, that is,

for every tangent vector s to Tp(M).

Proof. As in Section 1, let be a polar parametrization of Tp(M), withcorresponding geodesic polar mapping x = expp( ). (Henceforth we omitthe subscript from exp.)

At any point (u0, v0) in Tp(M), the tangent vectors u and v are anorthogonal basis for the tangent plane to Tp(M) at (u0, v0). As we saw inLemma 1.6, F = ·xu, xvÒ = 0, so the vectors

are also orthogonal.The proof of the lemma separates into three assertions.

(1) It suffices to show that the lengths of u and v are not decreasedby the differential map exp*.Proof. To show that this condition is sufficient, suppose that, in fact,||xu|| � || u|| and ||xv|| � || v||, and let s be an arbitrary tangent vector toTp(M) at (u0, v0). Now

exp exp ˜ ˜ exp ˜ exp ˜ .* * * *s x x x x x x( ) = +( ) = ( ) + ( ) = +a b a b a bu v u v u v

xxx

xx

x x x xu u v v= ( ) = ( )exp* ˜ exp* ãnd

xxxx

xx

expp( ) ( )* s s�

MM

MM


Then, since xu and xv are orthogonal,

(2) ||xu|| = || u||.

Proof. The definition of polar parametrization in Section 1 shows thatevaluated at an arbitrary (u0, v0), u is the velocity vector at u = u0 of theunit-speed radial line

in Tp(M). We know that exp carries this radial line to the unit-speedgeodesic u Æ x(u, v0) in M with initial velocity cosv0 e1 + sinv0 e2. Hence

(3) ||xv|| � || v||.

Proof. This time the comparison between in Tp(M) and x in M involvesthe Jacobi equation. The definition of polar parametrization in Section 1shows that

Thus, for the radial line v = v0,

(*)

Again exp carries this ray to a geodesic u Æ x(u, v0), along which

We know that g(u) = ||xv(u, v0)|| is uniquely determined by

Since K � 0, this implies

gg g gg g Kg¢( )¢ = ¢ + ¢¢ = ¢ -2 2 2 0� .

¢¢ + = ( ) = ¢( ) =g Kg g g0 0 0 0 1, , .

exp ˜ .* , ,x xv vu v u v u0 0 0( )( ) = ( ) ( )�

˜ .xv u v u u, 0 0( ) = ( )�

˜ sin cos .x e ev u v u v u v,( ) = - +1 2

x

x

˜ .x xu uu v u v0 0 0 01, ,( ) = = ( )

u u v u v u vÆ ( ) = +˜ cos sinx e e, 0 0 1 0 2

x

x

exp*

˜ ˜

.

s x x

x x

x x

s

( ) = +

= +

+

=

2 2 2

2 2 2 2

2 2 2 2

2

a b

a b

a b

u v

u v

u v�


But then gg¢ � 0 for u � 0, so

Using the initial conditions on g, we find (g¢2)(u) � (g¢2)(0) = 1 for u � 0.Since g¢ is initially positive, g¢(u) � u for all u � 0. Hence

Combining this with (*) completes the proof. ◆

If a map F is length nondecreasing in this sense, then evidently F*(v) = 0implies v = 0, so F is a regular mapping. Thus the exponential maps of com-plete K � 0 surfaces are local diffeomorphisms. As in earlier cases, the ques-tion now becomes, Are they covering maps? The length nondecreasingproperty provides the answer.

7.5 Lemma If an exponential map expp: Tp(M) Æ M of a completesurface is length nondecreasing, then curve segments in M can be lifted toany level in Tp(M).

Proof. Let a: [0 b] Æ M be a curve. If v in Tp(M) is carried to a (0) byexpp, we must prove that there is a curve segment : [0, b] Æ Tp(M) suchthat expp ( ) = a.

Assume that such a lift does not exist. Since expp is a local diffeomor-phism, there is a partial lift c defined on some initial segment Jc: 0 � t < cof [0, b]. Let d be the largest such partial lift. We will show that d can, infact, be extended past d—and this contradiction proves the lemma.

Let {ti} be an increasing sequence in Jd that approaches d. The liftedpoints (ti) in Tp(M) are in the set (Jd). Since expp does not decrease thelengths of vectors, it cannot decrease the length of curves; hence

Thus these points all lie in a closed disk D of radius L(a) centered at v. Astandard theorem of calculus asserts that every infinite sequence in a closeddisk in R2 contains a convergent subsequence. Since Tp(M) is isometric toR2, we can apply that result to the lifted points (ti) to get a subsequence{ (tn)} converging to some point w of Tp(M). Necessarily, tn Æ d.

Again, since expp is a local diffeomorphism, some neighborhood of wis carried diffeomorphically onto a neighborhood of a(d). But then theinverse l of this local map lifts a segment of a around a(d) into Tp(M),

aa

˜ ˜ exp ˜ .a a a a at L L L J Li d p d d( ) - ( )) ( )( ) ( ) < ( )v � � �

aa

aaa

aa

xv u v g u u u, 0 0( ) = ( ) ( )� � .

¢( )¢ = ¢ ¢¢ = - ¢ ( )g g g Kgg u2 2 2 0 0� � .


where (by the uniqueness of lifts) it agrees with for t < d and hence provides the sought-for extension of to values of t greater than d. ◆

Now we can prove the result mentioned earlier.

7.6 Theorem If M is a complete connected surface with K � 0, then forany point p in M, the exponential map expp: Tp(M) Æ M is a covering map.

Proof. We would like to use Theorem 4.12, which says that a local isom-etry with the geodesic lift property is a covering map. But the exponentialmaps in the theorem are generally not local isometries—in terms of thenatural Euclidean structure of Tp(M) ª R2. But as noted above, these expo-nential maps are regular, so temporarily assigning Tp(M) the pullbackmetric from M, as in Remark 1.3(2) of Chapter 7, makes them local isome-tries. The preceding lemma shows, in particular, that geodesics can belifted; thus Theorem 4.12 applies to complete the proof. ◆

7.7 Theorem (Hadamard) If M is a complete, simply connected surfacewith K � 0, then

(1) M is diffeomorphic to the Euclidean plane R2.(2) All nonconstant geodesics are one-to-one (so there are no geodesic

loops or closed geodesics).(3) Through any points p π q in M there is a unique (up to parametriza-

tion) geodesic.

Proof. (1) By the preceding theorem, exp is a covering map, and sinceM is simply connected, Theorem 4.8 says that exp is a diffeomorphism. Sofor any point p, we have diffeomorphisms

(2) If a nonconstant geodesic has, say, g (0) = g (1), then the exponen-tial map expg (0) sends both 0 and g ¢(0) π 0 to g (1). But exp is one-to-one.

(3) If geodesics a and b each pass through both p and q, then by repa-rametrization we can suppose

But expp sends a ¢(0) to a(1) and b¢(0) to b(1), hence

Since expp is one-to-one, a ¢(0) = b¢(0). Hence by the uniqueness of geo-desics, a = b. ◆

exp exp .p p¢( )( ) = ¢( )( )a b0 0

a b a b0 0 1 1( ) = ( ) = ( ) = ( ) =p q, .

R2 ª ( ) ªT M Mp .

aa


Thus M has the same differentiable structure as the Euclidean plane R2 andshares its most famous geometric property: “two points determine a line.”

Exercises

1. Prove that the exponential maps of a complete flat connected surface areRiemannian covering maps.

2. Show that under the hypotheses of Bonnet’s theorem, if area(M) = 4p/k,then M is isometric to a sphere S k of curvature k. (Hint: Compare the finalpart of the proof of Theorem 1.8, where less-than-or-equal successivelybecomes equality.)

3. Prove that a compact surface M is intrinsically bounded (has finite intrin-sic diameter) and if M is in R3 has finite Euclidean diameter.

4. If a complete connected surface has finite intrinsic diameter, show thatit is compact. (Hint: Use the scheme in the proof of Bonnet’s theorem.)

8.8 Summary

The global structure of a complete connected surface M can be described interms of geodesics and Gaussian curvature K. From a point p in M, run geo-desics out radially until (by the Hopf-Rinow theorem) they fill M. The onlygeometric difference from the Euclidean plane is the stretching of the polarcircles (orthogonal trajectories of the radial geodesics), and this stretching iscontrolled by K, via the Jacobi differential equation.

The topological difference from R2 is more serious, but the notion of cov-ering surface and the classification of compact surfaces lead to good results.In particular, we can give a reasonably complete account of all surfaces ofconstant curvature. When only inequalities are imposed on curvature, we stillget the powerful theorems of Bonnet (for K ≥ k > 0) and Hadamard (for K £ 0).

Generally speaking, the strictly geometric results of the last two chaptershold up well when geometric surfaces are replaced by higher-dimensionalRiemannian manifolds. Once the definitions are readjusted, most proofs areessentially the same. For example, the Hopf-Rinow theorem and the Bonnetand Hadamard theorems remain valid since their proofs depend only on whathappens on an individual geodesic. Dimension 2 has simplified certain

8.8 Summary 449

consistency proofs such as Theorems 2.1 and 3.2 in Chapter 7, but these canbe avoided entirely by more advanced methods.

The most striking changes come in the links between topology and geom-etry. In higher dimensions there is no classification of compact manifoldscomparable to Theorem 4.10 of this chapter. Where in dimension 2 only thesphere and projective space admitted metrics of positive curvature, now thereare many. The Euler characteristic c generalizes but no longer characterizes,and odd-dimensional compact manifolds all have c = 0. Nevertheless, theGauss-Bonnet theorem extends to higher (even) dimensions, and in this andother ways, curvature continues its strong influence over topology.


▼

▲Appendix

Computer Formulas

451

The computer commands most useful in this book are given in both theMathematica and Maple systems. More specialized commands appear in theanswers to several computer exercises. For each system, we assume a famil-iarity with how to access the system and type into it.

In recent versions of Mathematica, the core commands have generallyremained the same. By contrast, Maple has made several fundamentalchanges; however most older versions are still recognized. For both systems,users should be prepared to adjust for minor changes.

Mathematica

1. Fundamentals

Basic features of Mathematica are as follows:

(a) There are no prompts or termination symbols—except that a finalsemicolon suppresses display of the output. Input (new or old) is acti-vated by the command Shift-return (or Shift-enter), and the input andresulting output are numbered.

(b) Parentheses (. . .) for algebraic grouping, brackets [. . .] for argumentsof functions, and braces {. . .} for lists.

(c) Built-in commands typically spelled in full—with initials capitalized—and then compressed into a single word. Thus it is preferable for user-defined commands to avoid initial capitals.

(d) Multiplication indicated by either * or a blank space; exponents indi-cated by a caret, e.g., x^2. For an integer n only, nX = n*X, where Xis not an integer.

(e) Single equal sign for assignments, e.g., x = 2; colon-equal (:=) fordeferred assignments (evaluated only when needed); double equal signsfor mathematical equations, e.g., x + y == 1.

(f) Previous outputs are called up by either names assigned by the user or%n for the nth output.

(g) Exact values distinguished from decimal approximations (floatingpoint numbers). Conversion using N (for “numerical”). For example,E^2*Sin[Pi/3] returns ; then N[%] gives a decimal approximation.

(h) Substitution by slash-dot. For example, if expr is an expressioninvolving x, then expr/.xÆu^2 + 1 replaces x everywhere in theexpression by u2 + 1.

Mathematica has excellent error notification and online help. In particu-lar, for common terms, ?term will produce a description. Menu items giveformats for the built-in commands. The complete general reference book—exposition and examples—is The Mathematica Book [W]. For our purposes,the outstanding reference is Alfred Gray’s book [G].

> Some basic notation. Functions are given, for example, by

f[x_]:= x^3-2x+1 org[u_,v_]:= u*Cos[v]-u^2*Sin[v]

Here, as always, an underscore “_” following a letter (or string) makes it avariable. Thus the function f defined above can be evaluated at u or 3.14 ora2 + b2.> Basic calculus operations.Derivatives (including partial derivatives) by D[f[x],x] orD[g[u,v],v]Definite integrals by Integrate[f[x],{x,a,b}]. For numerical inte-gration, prefix an N thus: NIntegrate.

> Linear algebra. A vector is just an n-tuple, that is, a list v={v1,...,vn},whose entries can be numbers or expressions. Addition is given by v+w andscalar multiplication by juxtaposition, with sv=s{v1,...,vn} yielding{s*v1,...,s*vn}. The dot product is given by v.w and, for n = 3, thecross product is Cross[v,w].

Mathematica describes a matrix as a list of lists, the latter being its rows.For example, {{a,b},{c,d}} is a matrix and is treated as such in all contexts.

To make it look like , apply the command MatrixForm. The

determinant of a square matrix m is given by Det[m].

a bc d

ÊË

ˆ¯

e2 3 2

452 Appendix: Computer Formulas

The full power of the dot operator (.) appears only when matrices areinvolved. First, if p and q are properly sized matrices, then p.q is theirproduct. Next, if m is an m ¥ n matrix and v is an n-vector, then m.v givesthe usual operation of m on v. Taking m = n = 3 for example, if m1, m2,m3 are the rows of m and v={v1,v2,v3}, then Mathematica defines

m.v to be {m1.v,m2.v,m3.v}

This can be seen to be the result of m (in 3 ¥ 3 form) matrix-multiplying thecolumn-vector corresponding to v, with the resulting column-vector restatedas an n-tuple. In this sense, Mathematica obeys the “column-vector conven-tion” from the end of Section 3.1, which identifies n-tuples with n ¥ 1 matrices.

If A is any array—say, a vector or matrix—then for most commands,cmd[A] will apply the command cmd to each entry of A.

2. Curves

A curve in R3 can be described by giving its components as expressions in asingle variable. Example:

c[t_]:= {Cos[t],Sin[t],2t}

Then the vector derivative (i.e., velocity) is returned by D[c[t],t].

> Curves with parameters. For example, the curve c above can be generalizedto

helix[a_,b_][t_]:= {a*Cos[t],a*Sin[t],b*t}

Then helix[1,2]=c.

The following formulas, drawn from Theorem 4.3 of Chapter 2, illustrateaspects of vector calculus in Mathematica.

The curvature and torsion functions k and t of a curve c ª g are given by

kappa[c_][t_]:=Simplify[Cross[D[c[tt],tt],D[c[tt],{tt,2}]].Cross[D[c[tt],tt],D[c[tt],{tt,2}]]]^(1/2)/Simplify[D[c[tt],tt].D[c[tt],tt]]^(3/2)/.tt->t

(Note the description of second derivatives.) The use of the dummy variablett makes kappa[c] a real-valued function R Æ R. Otherwise, it would bemerely an expression in whichever single variable was used.

“Simplify” is the principal Mathematica simplification weapon; however, itcannot be expected to give ideal results in every case. (“FullSimplify” is more

Mathematica 453

powerful but slower.) Thus human intervention is often required, either to dohands-on simplification or to use further computer commands such as“Together” or “Factor” or trigonometric simplifications.

tau[c_][t_]:=Simplify[Det[{D[c[tt],tt],D[c[tt],{tt,2}],D[c[tt],{tt,3}]}]]/

Simplify[Cross[D[c[tt],tt],D[c[tt],{tt,2}]].Cross[D[c[tt],tt],D[c[tt],{tt,2}]]]/.tt->t

Here the determinant gives a triple scalar product.Note: The distinction between functions and mathematical expressions

is basic. Thus, with notation as above, tau applied to a curve, say,helix[1,2], is a real-valued function tau[helix[1,2]] whose valueon any variable or number s is tau[helix[1,2]][s].

The unit tangent, normal, and binormal vector fields T, N, B of a curvewith k > 0 are given by

tang[c_][t_]:=D[c[tt],tt]/Simplify[D[c[tt],tt].D[c[tt],tt]]^(1/2)/.tt->t

nor[c_][t_]:=Simplify[Cross[binor[c][t],tang[c][t]]]

binor[c_][t_]:=Simplify[Cross[D[c[tt],tt],D[c[tt],{tt,2}]]]/

Simplify[Factor[Cross[D[c[tt],tt],D[c[tt],{tt,2}]].

Cross[D[c[tt],tt],D[c[tt],{tt,2}]]]]^(1/2)/.tt->t

Here is how to preserve any such commands for future use: Type (or copy)them into a Mathematica notebook, say frenet, and use the Cell menu todesignate the cells containing them as initialization cells. When this notebookis saved, a choice will be offered letting you save, not only frenet, but alsoa new file frenet.m that contains only the commands. Then these can beread into later work by <<frenet.m

3. Surfaces

A coordinate patch, say x, is given by listing its components as expressionsin two variables. For example,

x[u_,v_]:= {u*Cos[v],u*Sin[v],2v}


> Parameters can be handled as above for curves. For example, the 2 in thisformula can be replaced by an arbitrary parameter using

helicoid[b_][u_,v_]:={u*Cos[v],u*Sin[v],b*v}

Then helicoid[2] gives the original x.

For a patch, the following commands return , and, , . We elect to represent our capital letters (E) by double lowercase letters(ee), since many capitals have special meaning for Mathematica (for example,E = 2.7183 . . .).

ee[x_][u_,v_]:=Simplify[D[x[uu,vv],uu].D[x[uu,vv],uu]]/.{uu->u,vv->v}

ff[x_][u_,v_]:=Simplify[D[x[uu,vv],uu].D[x[uu,vv],vv]]/.{uu->u,vv->v}

gg[x_][u_,v_]:=Simplify[D[x[uu,vv],vv].D[x[uu,vv],vv]]/.{uu->u,vv->v}

ww[x_][u_,v_]:=Simplify[Sqrt[ee[x][u,v]*gg[x][u,v]-ff[x][u,v]^2]]

The variant command, say www, in which Sqrt[...] is replaced by PowerExpand[Sqrt[...]] will often give decisively simpler squareroots. But one must check that its results are positive, since for example,PowerExpand[Sqrt[x^2]] yields x.

11[x_][u_,v_]:=Simplify[Det[{D[x[uu,vv],uu,uu],D[x[uu,vv],uu],D[x[uu,vv],vv]}]/ww[x][u,v]]/.{uu->u,vv->v}

The formulas for mm and nn are the same except that the double derivativeuu, uu is replaced by uu, vv and vv, vv, respectively.

> Gaussian curvature K. When the commands for E, F, G and , , havebeen read in, commands for K and H follow directly from Corollary 4.1 ofChapter 5 (see Exercise 18 of Section 5.4). However, the fastest way to findK for a given patch in R3 is by the following command, based on Exercise 20of Section 5.4. In it, “Module” creates an enclave in which temporary defin-itions can be made that let the final formula be expressed more simply.

gaussK[x_][u_,v_]:= Module[{xu,xv,xuu,xuv,xvv},xu=D[x[uu,vv],uu];xv= D[x[uu,vv],vv];

E F G W EG F, , , = - 2

Mathematica 455

xuu=D[x[uu,vv],uu,uu];xuv=D[x[uu,vv],uu,vv];xvv=D[x[uu,vv],vv,vv];Simplify[(Det[{xuu,xu,xv}]*Det[{xvv,xu,xv}]-Det[{xuv,xu,xv}]^2)/(xu.xu*xv.xv-(xu.xv)^2)^2]]/.{uu->u,vv->v}

As with other useful commands, this should be saved for future use.

4. Plots

There are four basic types: Plot and Plot3D plot the graphs of functionsof one and two variables respectively. Examples:

Plot[f[x]//Evaluate,{x,a,b}]Plot3D[g[x,y]//Evaluate,{x,a,b},{y,c,d}]

Here //Evaluate improves the speed of plotting.ParametricPlot plots the image of a parametrized curve in the plane

R2.ParametricPlot3D plots the image of a parametrized curve or patch.

For example, a parametrized curve c(t) in R3 is plotted for a � t � b by

ParametricPlot3D[c[t]//Evaluate,{t,a,b}]

and if x is an explicitly defined patch or parametrization, its image on therectangle 0 � u � 1, 0 � v � 2p is plotted by

ParametricPlot3D[x[u,v]//Evaluate,{u,0,1},{v,0,2Pi}]

Various refinements are available for plots. For example, if the end of thecommand above is altered to

...{v,0,2Pi},AspectRatio->Automatic]

then the same scale is imposed on height and width. Formally, the option“AspectRatio” has been reset from its default value. Various adjuncts to aplot can be also be changed. For example, the box surrounding the preced-ing plot is eliminated by Boxed->False. The plot can be made smootherby using PlotPoints->{m,n}, where the integers increase the defaultvalues governing smoothness in the u and v directions, respectively.

The options available for a command cmd are given, along with theirdefault values, by Options[cmd]. Then ?opt will describe a particularoption.

Previously drawn plots can be shown on the same page by

Show[plot1,plot2,plot3]


5. Differential Equations

Explicit solutions in terms of elementary functions are inherently rare, so wedescribe how to find and plot numerical solutions, which are all that is neededin many contexts. In the command for such a solution, Mathematica lumpsequations and initial conditions into a single list, then specifies the dependentvariables and the interval of the dependent variable.

Example: Solve numerically the differential equations


on the interval tmin � t � tmax. The format is

soln = NDSolve[{x9[t]==f[x[t],y[t],t],y9[t]==g[x[t],y[t],t],x[t0]==x0,y[t0]==y0},{x,y},{t,tmin,tmax}]

Note the double equal signs. Without the N for “numerical,” an exact solu-tion would be sought.NDSolve expresses x and y in terms of Interpolating Functions, data suf-

ficient for subsequent plots. If soln is an explicit result from the precedingcommand, the solution is plotted by

ParametricPlot[Evaluate[{x[t],y[t]}/.soln],{t,tmin,tmax}]

Here “/.” substitutes soln into the coordinates. Note the general equiva-lence: Evaluate[X] is the same as X//Evaluate.

Maple

1. Fundamentals

Basic features of Maple are as follows:

(a) Input is typed after a prompt and must be terminated by a semicolon—or colon, to suppress display of the output. We do not show thesebelow. Then press (or ).

(b) Parentheses used for algebraic grouping and arguments of functions;braces {. . .} for sets; brackets [. . .] for lists.

x t x y t y0 0 0 0( ) = ( ) =, ,

¢ = ( ) ¢ = ( )x f x y t y g x y t, , , , , ,

Maple 457

(c) Built-in commands are abbreviated, with multiword commands com-pressed into a single word; most are written in lower case.

(d) Multiplication always indicated by *, exponents by a caret, e.g., x^2.(e) Assignments indicated by colon-equal, e.g., x:=2; equations by single

equal, e.g., x+y=1.(f) Previous outputs are called up by names assigned by the user. (Naming

is important since input/outputs are not numbered.) Also, the percentsymbol (%) gives the immediately preceding output, and two of thesegive the one before that.

(g) Exact values distinguished from decimal approximations (floatingpoint numbers). Conversion is accomplished by the “evalf” command.For example, exp(2)*sin(Pi/3) returns ; then evalf(%) gives a decimal approximation.

(h) Substitution by the “subs” command. If expr is an expression involv-ing x, then subs(x=u^2+1, expr) replaces every x in the expres-sion by u2 + 1.

(i) If A is an array—say a matrix or vector—then to apply an operationF to each entry of A, use the command “map” thus: map(F,A).

Maple has a distinctive command “unapply” that converts mathematicalexpressions into functions. For example, if expr is an expression involving u and v, then unapply(expr,u,v) is the corresponding function ofu and v.

Many specialized Maple commands are collected in packages, which areloaded, for example, by with(plots). A list of the commands in thepackage appears unless output is suppressed. We rarely use packages otherthan plots and LinearAlgebra (which is replacing linalg).

Maple has reasonable error notification and excellent on-line help. Forcommon terms, ?term will produce a detailed description (no semicolonrequired).

The Maple Learning Guide is a good introduction to the most recentversion of Maple; it may be obtained from the website maplesoft.com. Ofcourse, there are a variety of more advanced books.

Some basic notations.

Functions can be produced by the arrow notation. Examples:

f:= x->x^3-2*x+1 org:= (u,v)->u*cos(v)-u^2*sin(v)

Derivatives (including partials):

diff(f(x),x) or diff(g(u,v),v)

e2 3 2


Definite integral:

int(f(x),x=a..b) orint(g(x,y),x=a..b,y=c..d)

If an explicit integral cannot be found, then evalf(%) gives a numericalresult. Direct numerical integration is given by evalf(Int(f(x),x=a..b)).

Linear algebra. Recent versions of Maple have changed considerably (thoughit still recognizes many old forms). Currently, its commands, whether new ornot, are often signalled by new names. Typically, the new command beginswith a capital letter and is not abbreviated. These changes are most evidentin the package LinearAlgebra that is replacing linalg.

Maple has always made a fundamental distinction between an n-tuple[v1,..,vn]—which is a list—and a vector, in any notation. The two typescannot directly interact. In the new version, vector is replaced by Vector(capital V).

Lists are the easiest to deal with. For instance, the usual sum of n-tuplesv=[v1,..,vn] and w=[w1,..,wn] is given by v+w, and scalar multi-plication of an n-tuple by a number s uses an asterisk, with s*v giving[s*v1,..,s*vn].

A matrix is produced by applying the command Matrix to a list whoseentries are lists, the latter being the rows of the matrix. Thus

Matrix([[a,b],[c,d]]).

With the package LinearAlgebra loaded, the determinant of a square matrixm is given by Determinant(m).

When an n ¥ n matrix C is considered as a linear transformation on Rn, itcannot directly attack [v1,..,vn] to give the image [w1,..,wn]. Thelist [v1,..,vn] must first be stood on end as Vector([v1,..,vn]),which is, in fact, an n ¥ 1 matrix. Now matrix multiplication is valid, and,with LinearAlgebra installed, Multiply(C,Vector([v1,..,vn]) isthe n ¥ 1 matrix that convert(%,list) turns into [w1,..,wn]. Thisidentification of an n-tuple with a column vector is just the “column vectorconvention” at the end of Section 3.1.

Since curves and surfaces are described in terms of lists, we can largely avoidthe list/Vector conflict by defining three basic vector operations directly interms of lists. First, note that the entries of a list p:=[p1,p2,. . .,pn]can be any expressions, and the ith entry is displayed by the command p[i].

a b

c dMaple

ÊËÁ

ˆ¯

is described by as

Maple 459

An operation applied to a list is automatically applied to each entry. (By con-trast, other arrays require the command map.)

Dot product: dot:=(p,q)–> simplify(p[1]*q[1]+

p[2]*q[2]+p[3]*q[3])

Cross product: cross:=(p,q)–> simplify

([p[2]*q[3]–p[3]*q[2],p[3]*q[1]–

p[1]*q[3],p[1]*q[2]–p[2]*q[1]])

Triple scalar product: tsp:=(p,q,r)–> dot(p,cross(q,r))

The built-in simplify above will reduce the number needed in later com-mands. Note that tsp(p,q,r) is just the determinant of the matrix withrows p,q,r, so reversal of any two entries gives (only) a sign change.

The three commands can be saved in Maple’s concise machine language by:

save dot,cross,tsp,99dotcrosstsp.m99

(Any name ending in “.m” will do as well.) These commands can then beintroduced into later sessions by

read 99dotcrosstsp.m99

(Formerly, save and read were expressed by save(cmd1,cmd2,’filename.m’) and read(’filename.m’), using backquotes.)

> Differential forms. The package difforms provides the essentials, includingthe exterior derivative operator d. The command defform is used to specifythe degree of the forms involved. For example, defforms(x=0,y=0) tellsMaple that x and y are 0-forms, that is, real-valued functions. Then thecommand d(x^2*sin(y)) yields 2x sin(y)d(x)+x2 cos(y)d(y).

2. Curves

A curve in R3 is described by giving its components as expressions in a singlevariable, for example, c:= t–>[3*cos(t),3*sin(t),2*t]. Then thevector derivative (i.e., velocity) of c is returned by diff(c(t),t), whichdifferentiates each component of the curve by t.

> Curves with parameters. For example, using the unapply command, thecurve c can be generalized to

helix:=(a,b)–> unapply([a*cos(t),a*sin(t),b*t]

Then helix(3,2) gives c as above.


> Frenet apparatus. We now show how the Frenet formulas in Theorem 4.3of Chapter 2 can be expressed in terms of Maple.

The curvature function k of a curve c ~ g is given by

kappa:= c –> unapply(simplify(dot(cross(diff(c(t),t),diff(c(t),t,t)),cross(diff(c(t),t),diff(c(t),t,t)))^(1/2)/dot(diff(c(t),t),diff(c(t),t))^(3/2)),t)

Here “unapply” makes kappa(c) a real-valued function on the domain ofc. Otherwise, it would merely be an expression in t and could not be evalu-ated on real numbers or other variables.

The command “simplify” is the principal Maple simplification weapon, butit not a panacea. It can be augmented by related commands such as “factor”or “expand.” Use ?simplify for information about these.

No set pattern of commands will give good results in every case, andhuman intervention is often required to get reasonable simplification.

The torsion function tau of a curve c is given by

tau := c -> unapply(simplify(tsp(diff(c(t),t),diff(c(t),t,t),diff(c(t),t,t,t))/factor(

dot(cross(diff(c(t),t),diff(c(t),t,t)),cross(diff(c(t),t),diff(c(t),t,t))))),t)

The distinction between functions and mathematical expressions is alwaysimportant. Thus, with notation as above, tau, applied to a curve, sayhelix(3,2), is a real-valued function whose value at a number or variables is given by tau(helix(3,2))(s).Maple has several varieties of scalar multiplication when LinearAlgebra isinstalled, however, since we are working with lists, s*v suffices.

The Frenet frame of a curve. The unit tangent, normal, and binormal vectorfields T, N, B of a curve c are given by

tang:=c->unapply(dot(diff(c(t),t),diff(c(t),t))^(-1/2)*diff(c(t),t),t)

nor:=c->unapply(cross(binor(c)(t),tang(c)(t)),t)

binor:=c->unapply(simplify(factor(dot(cross(diff(c(t),t),diff(c(t),t,t)),cross(diff(c(t),t),diff(c(t),t,t)))))^(-1/2)*cross(diff(c(t),t),diff(c(t),t,t)),t)

Maple 461

The presence of square roots in these formulas means that we cannot expectsimple results unless the curve itself is quite simple. However, individualvalues of the vector fields are usually readable.

Once the Frenet commands have been typed, they can be saved in a Mapledot-m file by

save kappa,tau,tang,nor,binor,99frenet.m99

and, as usual, these commands can be installed in later work by read99frenet.m99.

3. Surfaces

A coordinate patch, say x, in R3 is defined as a list-valued function whoseentries are expressions in two variables. For example,

x:=(u,v)–>[3*u*cos(v),3*u*sin(v),2*v]

Parameters in a patch can be handled as above for curves. For example, the3 and 2 in this formula can be replaced by an arbitrary parameters a and busing

helicoid:=(a,b)–> unapply([a*u*cos(v),a*u*sin(v),b*v],u,v)

Then helicoid(3,2) gives the original patch x.

The following commands, applied to a patch x, return E, F, G, W = EG- F 2, and , , . We elect to represent these capital letters (E) by doublelowercase letters (ee) since some capitals have special meaning for Maple (forexample, I = -1).

ee := x–> unapply(dot(diff(x(u,v),u),diff(x(u,v),u)),u,v)

ff := x–> unapply(dot(diff(x(u,v),u),diff(x(u,v),v)),u,v)

gg := x–> unapply(dot(diff(x(u,v),v),diff(x(u,v),v)),u,v)

(Recall that simplify is built into the dot command, defined earlier.)

ww := x–> unapply(simplify(ee(x)(u,v)*gg(x)(u,v)–ff(x)(u,v)^2)^(1/2),u,v)


ll := x–> unapply(tsp(diff(x(u,v),u,u),diff(x(u,v),u),diff(c(u,v),v))/ww(x)(u,v),u,v)

The formulas for mm and nn are the same, except that the double derivativeu,u is replaced by u,v and v,v, respectively.

As before, these commands can be saved by

save ee,ff,gg,ww,ll,mm,nn,99efgwlmn.m99

> Gaussian and mean curvature. When the commands above for E, F, G and, , have been read in, commands for K and H follow immediately fromCorollary 4.1 of Chapter 5. However, a faster way to find K for a given patchin R3 is to use the following command, based on Exercise 20 of Section 5.4.In it, proc, for “procedure”, begins an enclave—terminated by end proc—within which definitions can be made that do not escape to the outside.These temporary definitions allow the final formula to be expressed moreconcisely.

gaussK := proc(x)local xu,xv,xuu,xuv,xvv;xu := diff(x(u,v),u);xv := diff(x(u,v),v);xuu := diff(x(u,v),u,u);xuv := diff(x(u,v),u,v);xvv := diff(x(u,v),v,v);

unapply(simplify(factor(tsp(xuu,xu,xv)*tsp(xvv,xu,xv)–tsp(xuv,xu,xv)^2)/

(dot(xu,xu)*dot(xv,xv)–dot(xu,xv)^2)^2),u,v)end proc

Here tsp is the triple scalar product, defined earlier. As usual, gaussK canbe saved for future use.

4. Plots

Maple has three basic plot commands.

(1) The command plot has two uses:(i) Graphs. If f is a real-valued function defined on a � t � b, then

plot(f(t),t=a..b) draws its graph.(ii) Parametric plots. If g is another such function, then the curve

with c(t) = [f(t),g(t)] is plotted in R2 by plot(c(t),t=a..b). Alternatively, plot([f(t),g(t)],t=a..b)gives the same result.

Maple 463

Plots can be modified by options, thus: plot([c(t),t=a..b],<option>), where, for example, the option numpoints=200 wouldincrease the smoothness of the plot, and scaling=constrainedimposes the same scale on the axes. Use ?plot[options] to get manyothers.

(2) The command plot3d also has two uses. Let D be a region a � u �b, c � v � d in R2. Then(i) Graphs. If ƒ is a real-valued function defined on D, its graph is

plotted by plot3d(f(u,v),u=a..b,v=c..d).(ii) Parametric plots. If x:D Æ R3 is a list-valued patch or parame-

trization, its image is plotted by plot3d(x(u,v),u=a..b,v=c..d).

Again, ?plot3d describes a number of ways to specify plot style.(3) Parametrized curves in R3 are plotted using the command “spacecurve”

from the plots package. As an example: spacecurve(c(t),t=–2..4)

To show more than one plot on the same page, each plot should benamed, say, A:= plot3d(x(u,v),u=0..1,v=0..Pi): with termi-nal colon to avoid a flood of numbers. Then use “display” from the plotspackage: display([A,B,C]).

5. Differential Equations

Explicit solutions in terms of elementary functions are rare, so we describehow to find and plot numerical solutions, which are just as useful in manycontexts. In the command for a numerical solution, Maple lumps equationsand initial conditions into a single set, then gives the dependent variables (asfollows).

For example, suppose we want to solve numerically the equations


on the interval a � t � b. The format is

numsol:= dsolve({diff(x(t),t)=f(x(t),y(t),t),diff(y(t),t)=g(x(t),y(t),t),x(t0)=x0,y(t0)=y0},{x(t),y(t)},type=numeric)

x t x y t y0 0 0 0( ) = ( ) =,

¢ = ( ) ¢ = ( )x f x y t y g x y t, , , , ,


This solution is plotted by a command from the plots package:

odeplot(numsol,[x(t),y(t)],a..b)

Only now is the domain a � t � b of the solution specified.

Computer Exercises

Chapter 2: 2.2/9, 2.4/11, 14, 15, 19, 20, 2.7/7Chapter 3: 3.2/5, 3.5/4, 5, 9, 10Chapter 4: 4.2/5, 6, 11, 4.3/6, 11, 4.6/6, 4.8/10Chapter 5: 5.4/16, 18–21, 5.6/16, 18, 5.7/8, 9Chapter 6: 6.5/6, 6.8/11, 13Chapter 7: 7.2/13, 7.5/9–12, 7.7/12, 13Chapter 8: 8.1/8

Computer Exercises 465

▼

▲Bibliography

467

[dC] do Carmo, M. P., Differential Geometry of Curves and Surfaces, Prentice-Hall,1976.

[G] Gray, Alfred., Modern Differential Geometry of Curves and Surfaces, 2d ed., CRCPress, 1998.

[Ma] Massey, W. S., Algebraic Topology: An Introduction, Springer-Verlag, 1987.[Mi] Milnor, John W., Morse Theory, Princeton University Press, 1963.[Mu] Munkres, J. R., Topology, 2d ed., Prentice-Hall, 2000.[ST] Singer, I. M., and J. A. Thorpe, Lectures on Elementary Topology and Geometry,

Springer-Verlag, 1976.[S] Struik, D. J., Lectures on Classical Differential Geometry, 2d ed., Addison-Wesley,

1961. Reprint, Dover, 1988.[W] Wolfram, S., The Mathematica Book (various editions), Wolfram Media.

The book by do Carmo is a clearly written exposition of differential geometry with a viewpoint similar to this one, but at a more advanced level.Gray’s book is recommended for readers interested in the use of computers,especially for differential geometry. Both these books have extensive bibliographies.

▼

▲Answers to Odd-Numbered Exercises

468

These answers are not complete; and in some cases where a proof is required,we give only a hint.

Chapter 1

Section 1.1

1. (a) x2y3 sin2 z.(c) 2x2y cosz.

3. (b) 2xeh cos(eh), h = x2 + y2 + z2.

Section 1.2

1. (a) -6U1(p) + U2(p) - 9U3(p).3. (a) V = (2z2/7)U1 - (xy/7)U3.

(c) V = xU1 + 2yU2 + xy2U3.5. (b) Use Cramer’s rule.

Section 1.3

1. (a) 0.(b) 7 · 27.(c) 2e2.

3. (a) y3.(c) yz2(y2z - 3x2).(e) 2x(y4 - 3z5).

5. Use Exercise 4.

Section 1.4

1.

3.5. The lines meet at (11, 7, 3).7. vp = (1, 0, 1)p at p = (0, 1, 0).

Section 1.5

1. (a) 4.(b) -4.(c) -2.

3. Use Exercise 2 and f((1/x)V + (1/y)W) = f(V )/x + f(W )/y.5. (b) (x dy - y dx)/(x2 + y2).7. (a) dx - dz.

(b) not a 1-form.(c) zdx + xdy.

9. ±(0, 1, 1/2).11. (a) Consider the Taylor series for t Æ f(p + tv).

(b) Exact: -.420, approximate: -.500.

Section 1.6

1. (a) f Ÿy = yz cosz dx dy - sinz dx dz - cosz dy dz.(b) df = -z dx dy - y dx dz, since d(dz) = 0.

7. Apply this definition to the formula following Definition 6.3.9. For the alternation rule, set f = y, g = x.

Section 1.7

1. (c) (0, 0), (1, 0).3. F* (v) = F(p + tv)¢(0) = 2(p1v1 - p2v2, v1p2 + v2p1) at F(p).5. F* (vp) = F(p + tv)¢(0) = (F(p) + tF(v))¢(0) = F(v)F(p).

b s s s s( ) = + - -( )1 1 2 12, , .

¢( ) = -( ) ( )a p 2 1 0 1 2 1 1 2, , at , , .

Chapter 1 Answers 469

7. Using Lemma 4.6 gives vp[g(F)] = (d/dt)|0 g(F(p + tv)) = F(p + tv)¢(0)[g]= F* (vp)[g].

9. (a) GF = (g1(f1, f2), g2(f1, f2)).(b) (GF)* (a ¢(0)) = (GF(a))¢(0) = G* (F(a)¢(0)) = G*F* (a ¢(0)).(c) F-1 is one-to-one and onto. To show it is regular, start from

F(F -1) = I, the identity map. Hence F* (F -1)* = I* = identity mapon tangent vectors. So (F -1)* cannot carry a nonzero vector to zero.

Chapter 2

Section 2.1

1. (a) -4.(b) (6, -2, 2).

(c)

(d) .

(e) .5. If v ¥ w = 0, then u • v ¥ w = 0 for all u; use Exercise 4.7. v2 = v - (v • u) u.

Section 2.2

1. (b) s(t) = 2t + t3/3.

3. .

7. For (ii), |h¢| = -h¢ � 0, so the change of variables formula in an

integral gives

9. L(a) ª 12.9153 < 14.1438 ª L(b).

Section 2.3

1. k = 1, t = 0, B = -(3, 0, 4)/5, center (0, 1, 0), radius 1.7. (a) 1 = ||a(h)¢|| = ||a ¢(h)h¢|| = |h¢|, hence h¢ = ±1.

(b) Let e = ±1. Then = a(h) implies T = a ¢(h)h¢ = eT(h). Hence = k (h)N(h), and so on.Nk

a

dt dt Lb

a

a

b

a a a- ¢ = ¢ = ( )Ú Ú .

L h h ds h h ds h dsc

d

c

d

c

d

a a a a( )( ) = ( ) = ¢( ) - ¢( ) = - ¢ ¢ =Ú Ú Ú¢

b s s s s( ) = + ( )( -1 2 2 22 1/ / ,sinh /,

-2 15/

2 11

1 2 1 6 1 0 3 10, , , ,-( ) -( )/ , / .

470 Answers to Odd-Numbered Exercises

9. For the rectifying plane. From the formula for in the text, delete b (0)and the N0 term. The remaining terms give the same general shape asthe curve (s, ±s3).

11. (b) First differentiate B = ; consider the two ± cases and differentiateagain.

Section 2.4

1. (a) Let f = t2 + 2. Then k = t = 2/f 2 and B = (t2, -2t, 2)/f.(c) All the limits are natural unit vectors, ±(1, 0, 0), . . .

3. (a) N = (0, -1, 0), t(0) = 3/4.7. (a) (g (t) - a (t0)) • u = 0.

(b) g has constant speed, so use Exercise 5.9. Evidently, a is a cylindrical helix. By Exercise 7 its cross-sectional curve

g is a plane curve with constant curvature, hence g lies in a circle.11. (c) (Mathematica):

helix[a_,b_][t_]:={a*Cos[t],a*Sin[t],b*t}ParametricPlot3D[{helix[2,1][t],helix[–.5,1][t]}//Evaluate,{t,0,6Pi}](Maple): With the plots package installed,helix:=(a,b)–>[a*cos(t),a*sin(t),b*t]spacecurve({helix(2,1)(t),helix(–.5,1)(t)},t=0..6*Pi,numpoints=100)Recall that we do not show Maple’s mandatory terminal semicolon.

13. (b) lt(s) = a (t) + s(a ¢(t) • a ¢(t)/a ≤(t) • J(a ¢(t)) J(a ¢(t)) for 0 � s � 1.(c) For a unit speed, lt(s) = a + s(1/ )N. Hence dlt/ds = (1/ )N (inde-

pendent of s). Evidently this is normal to a at a (t). Since a* =a + (1/ )N, we get (a*)¢ = T + (1/ )¢N - T = (1/ )¢N, in agree-ment with dlt/ds at a* (1).

15. (a) For the rectifying plane (orthogonal to N):(Mathematica):viewN[a_,eps_]:=ParametricPlot[{(a[t]–a[0]).tang[a][0],(a[t]–a[0]).binor[a][0]}//Evaluate,{t,–eps,eps}]

(Maple)viewN:=(a,eps)–>plot([dot((a(t)–a(0)),tang(a)(0)),dot((a(t)–a(0)),binor(a)(0)),t=–eps..eps])

kkk

kk

B

b


(b) (iii) For all curves with t(0) π 0 there are essentially only two cases,depending on the sign of t.

17. (a) .(b) •.

(c) .(d) 2p (see Exercise 18).

19. (c) For a suitable n, let tn be t with new z-component (1/n) sin3t. Here= k, and in the notation of Exercise 12, .

21. Use Theorem 4.6. By hand computation (easy, if k and t are first foundby computer), we get t/k = (3ac/2b2)(P/Q)3/2, where

Thus t /k is constant if and only if 4b2 = 9a2c2/b2, that is, 3ac = ±2b2.(Hence t /k = ±1).

Section 2.5

1. (a) 2U1(p) - U2(p).(b) U1(p) + 2U2(p) + 4U3(p).

5.

Section 2.6

1. Show that V · = 0, and use Lemma 1.8.3. For instance, E2 = -sinzU2 + coszU3 and E3 = E1 ¥ E2.

Section 2.7

1. w12 = 0,3. w12 = -df, w13 = cos f df, w23 = sin f df.5. By Corollary 5.4(3),7. (Mathematica):

(a) connform[A_]:=Simplify[Dt[A].Transpose[A]](b) In A, write q for J and f for j. Then in MatrixForm

[connform[A]], read Dt[q] as dq.

(Maple): Install the packages LinearAlgebra and difforms. With q and f as above, write defform(q=0,f=0) to identify them as real-valued functions.(a) connform:=A–>simplify(Multiply(map(d,A),

Transpose(A)))

— ( ) = [ ] + —V i i i i i V if E V f E f ES S

w w13 23 2= = df / .

W

— = Â ¢( )[ ] = Â( ) ( )( )( ) = ( )¢ ( )¢( )a aa at i i i iW t w U d dt w t U W t/ .

P c t b t a Q c t a c b t a= + + = + ( ) +9 4 9 92 4 2 2 2 2 4 2 2 2 2 2and

ds dt x y= ¢ + ¢2 2k

p / 2

p / 2


Section 2.8

3. (a) Compute q = Adx, as in the text. (A was found in Section 7.)(b) For example, E1[r] = dr[E1] = q1(E1) = 1.(c) Use the appropriate form of the chain rule.

Chapter 3

Section 3.1

3. (Ta)-1 = T-a, and since C is orthogonal, C-1 = tC. Thus F-1 = (TaC)-1 =C -1 (Ta)-1 = tCT-a. By Exercise 1, this equals TtC (-a)

tC = T-tC(a)tC.

5. (b) Using Exercise 3 we find 7. Use Exercises 2 and 3.9. (a) For J such that C(1, 0) = (cosJ, sinJ), C has matrix

(b) O(1) consists of +1 and -1, so F(t) = a ± t for any number a.

Section 3.2

1. T(vp) = vT(p).3. The middle row of C is (-2, 1, 2)/3, and T is translation by

5. (Mathematica):Let ame={e1,e2,e3} and amf={f1,f2,f3} be the attitudematrices of the frames in Exercise 3.(b) Set cc:=Simplify[Transpose[amf].ame] Then Sim-

plify[cc.e1] is f1, etc.(Maple):Install the package LinearAlgebra, and let ame=Matrix([e1,e2,e3]) and amf=Matrix([f1,f2,f3]) be the attitude matrices ofthe frames in Exercise 3.(b) Set cc:=simplify(Multiply(Transpose(amf), ame)).

Then simplify(Multiply(cc,Vector(e1))) isVector(f1), etc.

3 1 2 2 3, 4/3,- -( )/

cos sin

sin cos.

J JJ J

m

±ÊËÁ

ˆ¯

F - ( ) = -( )1 5 2 2 3 2p , ,


Section 3.3

1. If the orthogonal parts of F and G are A and B, then by Exercise 2 ofSection 1, sgn(FG) = det AB = (detA)(det B) = det BA = sgn(GF ).Then +1 = sgnI = sgn(FF -1) = sgn(F )sgn(F -1).

5. C is rotation through angle p/2 about the axis given by a.

Section 3.4

1. (b) By definition, s(s) is the point canonically corresponding to T(s);hence by Exercise 1 of Section 2, C(s ) corresponds to F* (T ), theunit tangent of F(b ).

3. Translate each triangle so that its new first vertex is at the origin. A sketchwill show that the required C is orientation-reversing, and we find C =

5. For a tangent vector v at p,

Section 3.5

3. Take a = 2, b = ±2.5. Yes, since c has constant speed, curvature, and torsion.

7.

9. For simplicity, assume a � 0 � b; then:(Mathematica):(a) kdetc[f_,a_,b_]:=

NDSolve[{x’[s]==Cos[phi[s]],y’[s]==Sin[phi[s]],phi’[s]==f[s],x[0]==0,y[0]==0,phi[0]==0},{x,y,phi},{s,a,b}]

(b) draw[f_,a_,b_]:=ParametricPlot[Evaluate[{x[s],y[s]}/.kdetc[f,a,b]],{s,a,b},AspectRatio–>Automatic]

(Maple):(a) kdetc:=f–>dsolve({diff(x(s),s)=cos(phi(s)),

diff(y(s),s)=sin(phi(s)),diff(phi(s),s)=f(s),x(0)=0,y(0)=0,phi(0)=0},{x(s),y(s),phi(s)},type=numeric)

b j j js s ds s ds s f s ds( ) = ( ) ( )( ) ( ) = ( )ÚÚ Úcos sin, , where

F W F W t W F tC Wv F v* *( ) ( .*

—( ) = +( )¢ ( ) = ( ) + ( )( )¢ ( ) = — ( )p v p v0 0

-ÊËÁ

ˆ¯

3 5 4 5

4 5 3 5

/ /

/ /


(b) Install plots. Define draw:=(f,a,b)–>odeplot(kdetc(f),[x(s),y(s)],a..b,scaling=constrained).

Chapter 4

Section 4.1

1. (a) The vertex.(b) All points on the circle x2 + y2 = 1.(c) All points on the z axis.

5. (b) c π -1.9. Use Exercise 7.

11. q is in F(M) if and only if F-1(q) is in M, that is, g(F-1(q)) = c. Use theHint to apply Theorem 1.4.

Section 4.2

1. (c) The Monge patch x(u, v) = (u, v, u2 + v2) covers the entire surface;a parametrization based on Example 2.4 omits the point (0, 0, 0).

3. xu ¥ xv = vd ¢ (u) ¥ d(u).5. (a) EG - F 2 = b2 + u2 is never zero.

(b) Helices and straight lines (rulings).(c) H: xsin(z/b) - ycos(z/b).(d) For x as given:

(Mathematica): ParametricPlot3D[x[u,v]//Evaluate, {u,–1,1},{v,0,2Pi}]

(Maple): plot3d(x(u,v),u=–1..1,v=0..2*Pi)7. (b) x(u, v) = (cosu - vsinu, sinu + vcosu, v).9. In all cases, (i) check that the three partial derivatives of the defining

function g are never zero simultaneously on M: g = 1 (Theorem 1.4),and (ii) First, check that the components of x satisfy the equation g = 1.

11. (c) x ± (u, v) = (acosu, bsinu, 0) ± v(-asinu, bcosu, c).(d) (Mathematica):

xplus[u_,v_]:={1.5*(Cos[u]–v*Sin[u]),Sin[u]+v*Cos[u],2v}

ParametricPlot3D[xplus[u,v]//Evaluate,{u,0,2Pi},{v,–1,1}]

(Maple): xplus:=(u,v)Æ[1.5*(cos(u)–v*sin(u)),sin(u)+v*cos(u),2*v]plot3d(xplus(u,v),u=0..2*Pi,v=–1..1)


Section 4.3

1. (a) r2 cos2 v.(b) r2(1 - 2cos2 v cosu sinu).

3. (a) and are the Euclidean coordinate functions of x-1y.(b) Express y = x( , ) in terms of Euclidean coordinates, and

differentiate.5. (a) M is given by g = z - f(x, y) = 0, with —g = (-fx, -fy, 1), and v is

tangent to M at p if and only if v • —g(p) = 0.7. —g = (-y, -x, 1) is a normal vector field; V is a tangent vector field if

and only if V • —g = 0, for example, V = (0, 1, x).9. (a) p(M) consists of all points r such that (r - p) • z = 0; hence vp is

in Tp(M) (that is, v • z = 0) if and only if p + v is in p(M).11. (a) If a/b = m/n for integers m, n, consider Dt = 2p m/a = 2p n/b.

(b) Assume a(s) = a(t) for s π t, so x(as, bs) = x(at, bt). Equality forz components and for x2 + y2 implies as - at = 2pm and bs -bt = 2pn for some integers m, n. Thus a/b = m/n, a contradiction.

Section 4.4

1.

3. If a is a curve with initial velocity v at p, then

5. On the overlap of Ui and Uj, dfi - dfj = d(fi - fj) = 0.

7. (b) .

Section 4.5

1. If x: D Æ M is a patch, then F(x): D Æ N is (by Theorem 3.2) a dif-ferentiable mapping. Hence y-1Fx is differentiable for any patch y in N.

3. If and are patches in M and N, respectively, note that -1F =( -1y)(x-1 ) is differentiable, being a composition of differentiable functions.

5. By Exercise 1, A is differentiable. Since A2 = I, A-1 = A, so A is a dif-feomorphism. For A*, consider its effect on a curve t Æ cos tp + sin tuin S .

7. Theorem 5.4.

xyxyyx

du uu

uuuu u˜ ˜

˜x x

x( ) = [ ] =∂ ( )( )

∂=

∂∂

= 1

v p vp pg f gf g f f g f f( )[ ] = ( )¢( ) = ¢( )( )( )¢( ) = ¢ ( )( ) [ ]a a a0 0 0 .

df fd u vf f= Ÿ +( )( )x x, .

d ff

uf

vf

u vu v v u v uf f f f f( )( ) =∂ ( )( )

∂( ) -

∂ ( )( )∂

( ) + ( ) ∂∂

( ) -∂∂

( )ÈÎÍ

˘˚

x xx

xx

x x x x,

TT

vuvu


9. (a) Use Exercise 8.(b) F* (axu + bxv) = ayu + byv implies linearity.

11. M is diffeomorphic to a torus if the profile curve a of M is closed, andto a cylinder if a is one-to-one. With parametrizations as suggested,F(x(u, v)) = y(u, v) is a diffeomorphism.

13. (a) If p in M, there is a q in such that p = G(q). By consistency,F(p) = (q) is a valid definition. G is regular, hence locally has differentiable inverse mappings. Thus, locally F = G-1 so F is differentiable.

(b) If F(p1) = F(p2), then for q1, q2 in such that G(q1) = p1, G(q2) =p2, we have F(G(q1) = F(G(q2)). Thus (q1) = (q2). Then thehypothesis gives G(q1) = G(q2), that is, p1 = p2.

Section 4.6

3. (b) Use Theorem 6.2.5. (a) Let r(t) = ||a(t)||. Then let f = U1 • a/||a || and g = U2 • a/||a ||. Apply

Exercise 12 of Section 2.1 to get J.(b) J(a) and J(b) measure the same angle; hence they differ by some

integer multiple of 2p.(c) Use Exercise 1 to evaluate y on the polar expression for a in (a).

(d)

7. (a) Since (F*(f))(a¢) = f((F*)(a ¢)) = f(F(a)¢), we get

9. (a) 2pm, (b) 2pn.13. The text shows that if f is the dual of V, then The dual

of curl V is df, and dA ª W du dv. It follows that

Section 4.7

1. (a) Connected, not compact.(c) Not connected and not compact.(e) Connected and compact.

3. If v is nonvanishing on N, show that F*(v) is nonvanishing on M.5. (a) All—by Definition 7.1.

(b) Sphere, torus—by Lemma 7.2.

U V dA VW

W du dv du dvu vu v• • .curl curl ,=

¥= ( )x x

x xf

V ds• .Ú Ú= f

F F dta

b

F* .f f a f

a a( ) = ( )¢Ê

Ëˆ¯ =Ú Ú Ú ( )

det

•.

a aa a

, ¢( )=

¢ ¢+( ) =

¢ - ¢+

f g

f gf g

fg gff g

2 22 2

FFM

FF

M


(c) All—by Proposition 7.5(d) Plane, sphere (see text).

9. (c) If M is connected, then path-connectedness (Definition 7.1) followsusing parts (a) and (b). If M is path-connected, let U and M - U

be open sets of M such that U contains a point p.Assume that M - U contains a point q. There is a curve segment

a:[a, b] Æ M from p to q. Since a is continuous, a-1(U ) and a-1

(M - U ) are disjoint open sets filling [a, b]. This contradicts thestated connectedness of [a, b].

11. Fix q in M - R; then by the Hausdorff axiom, for each p in R, thereare disjoint neighborhoods Up of p and Uq,p of q. By compactness, afinite number of the neighborhoods Up cover R. Then the intersectionof the corresponding neighborhoods Up,q is a neighborhood of q thatdoes not meet R.

Section 4.8

1. If M is orientable it has a nonvanishing 2-form m. Then f(t) = m(a ¢(t),Y(t)) is a differentiable function on [a, b]. By (ii), f(a) f(b) < 0; hence f issomewhere zero on a < t < b. This contradicts (i).

5. (a) The function p Æ d(0, p) is continuous on M, hence takes on amaximum.

7. (i) Since M is nonorientable, there is a reversing loop (as in the hint)at some point q. Fix Uq. Then every point Up in M can be connectedto Uq by a curve in M. Proof: Move Up along a curve from p to q.If the result is -Uq, move it around the reversing loop.

9. (b) B - b is diffeomorphic to an ordinary band.11. (a) Recall that a neighborhood in a surface is the image under a coor-

dinate patch of a neighborhood in R2. Evidently every neighbor-hood x(U ) of 0 meets every neighborhood y(V ) of 0*.

(b) The sequence {(1/n, 0)} converges to 0 when expressed in terms ofx, and to 0* in terms of y.

(c) Relative to x and y, the coordinate form of F is the identity map.13. (a) In terms of the natural coordinates, a¢(t) = V(a(t)) becomes

(b) The differential equations are u¢ = -u2, v¢ = uv, and the initial conditions are u(0) = 1, v(0) = -1. The first differential equation integrates to 1/u = t + A. But u(0) = 1, so u = 1/(t + 1). Thus weget v¢ = v/(t + 1), which integrates to v = B(t + 1). Then v(0) =-1 implies v = -(t + 1).

¢ + ¢ = ( ) + ( )u U v U f u v U f u vU1 2 1 1 2 2, , .


15. Smooth overlap follows from the identity

Chapter 5

Section 5.1

1. Use Method 1 in the text.3. (a) 2.

(c) 1.5. Meridians go to meridians (great circles through the poles), parallels to

parallels—except for the top and bottom circles of the torus.7. Use Method 1 and the definition of tangent map in Chapter 1.

Section 5.2

1. (b) If e1, , then S(e1) = e1 and S(e2) = -e2.

Section 5.3

1. k1k2 � 0 and k1 = k2 imply k1 = k2 = 0.5. (b) K > 0: an ellipse on one side and no points on the other. K < 0: the

two branches of a hyperbola. K = 0, nonplanar: two parallel lineson one side, no points on the other.

7. (a) If a is a curve with initial velocity v at p, then F*(v) = F(a)¢(0) =(a + eUa)¢(0) = v - eS(v) at F(p).

Section 5.4

1. W = r2 cosv > 0, U = x/r, K = 1/r2, H = -1/r.5. Use a ¢ = a¢1xu + a¢2xv to find speed.7. K = -36r2/(1 + 9r4)2.9. Expand S(v) ¥ v. This vector is zero if and only if its dot product with

xu ¥ xv is zero. Use the Lagrange identity (Exercise 6 of Section 3).11. k(u) = S(v) • v/v • v. Substitute v = v1xu + v2xv.15. (a) K is negative except at the origin, but this is a planar point, hence

an umbilic with k = 0.

e u u2 1 2 2= ±( ) /

x y x y xx yy¥( ) ¥( ) = ( ) ¥ ( )- - -1 1 1 .



(b) The hint leads to . These two umbilics reduce to one for the paraboloid of rotation, a = b, where(by symmetry) we expect 0 to be umbilic.

17. (b) Since k < B, if e < 1/B, then xu ¥ xv π 0.(c) S(xu) ¥ S(xv) = -kcosvT ¥ xv/e.

19. (Mathematica):(b) hyperboloid[a_,b_,e_][u_,v_]:=

{u,v,u^2/a^2+e*v^2/b^2}(c) monkeypolar[r_,q_]:=monkey[r*Cos[q],

r*Sin[q]](Maple):(b) hyperboloid:=(a,b,e)–>unapply([u,v,u^2/a^2+

e*v^2/b^2],u,v)(c) monkeypolar:=(r,q)–>monkey(r*cos(q),

r*sin(q))21. Maple has a built-in tube command in the plots package. For (c), with

t defined as in the exercise referred to, the tube is plotted by tubeplot(t(t),t=0..2*Pi,radius=0.5)(Mathematica):(a) With the commands for unit normal and binormal installed (see

Appendix), a tube formula istube[c_,r_][t_,phi_]:=c[t]+r*(Cos[phi]*nor[c][t]+Sin[phi]*binor[c][t])This is plotted—in (b), for example—byParametricPlot3D[tube[helix,1/2][t,phi]//Evaluate,{t,0,4Pi},{phi,0,2Pi},PlotPoints–>{40,20},Axes–>None,Boxed–>False]

(c) If the general approach in (a) is slow in this case, a faster way is tocopy the outputs of binor[t][t,phi] and nor[t][t,phi]into an explicit definition of the tube function of t.

Section 5.5

3. (a) The critical points of K are those of h. They occur at the interceptsof M with the coordinate axes.

(b) For the ellipsoid, c2/(a2b2) � K � a2/(b2c2). (Note again the effect ofa = b = c.)

5. (c) Use Z = grad(ez cosx - cosy) and W = Z ¥ V. Then —VZ ¥W + V ¥ —W Z = 0 and V • —VZ ¥ —WZ = -e2z.

0 2 42 2 2 2, ,± ( ) - -( )( )b a b a b


7. (a) Use (b) The tangency condition for a vector v at p is

Section 5.6

3. Use Remark 6.10.5. Since U • V is constant, U¢ • V + U • V¢ = 0. If a is principal in M, then

using Lemma 6.2, U¢ • V = 0, hence V • U¢ = 0. Continue as for Lemma6.3.

7. S(T) = -U¢; hence by orthonormal expansion, U¢ = -S(T) • T T -S(T) • V V. Continue as in the proof of the Frenet formulas.

11. (a) Set s = a + fd. Then f is determined using the equation s ¢ • d¢ = 0.(b) d ¢ ^ d, a ¢ implies that a ¢ ¥ d and d ¢ are collinear. Then a ¢ ¥ d =

pd ¢. Hence xu ¥ xv = pd ¢ + vd ¢ ¥ d, so W 2 = (p2 + v2)d ¢ • d ¢.Now use Exercise 12 of Section 4.

(c) On each ruling, K has a unique minimum point; the striction curvemeets the ruling at this point.

13. (a) Since s (u + e) - s (u) ª es¢(u), the Hint gives de = es¢ (u) • d(u) ¥d ¢(u)/||d (u) ¥ d ¢(u)||. However ||d (u) ¥ d ¢(u)||e ª ||d (u) ¥ d(u + e)||= sinJe ª Je. Since ||d(u) ¥ d ¢(u)||2 = d ¢ • d ¢, we see that limeÆ0de/Je

= s ¢ • d ¥ d¢/d ¢ • d ¢ = p.15. Compute E, F, G and , , . (Computer formulas for these are given

in the Appendix.) Then EG - F 2 π 0 proves (a), and F = = 0 proves(b).

17. (a) K = -h¢2J ¢2/W 4, H = u(h¢J ≤ -J ¢h≤)/(2W3), where W2 = h¢2 + u2J ¢2.(b) d ¥ d ¢ = J ¢U3. Since K is a minimum when u = 0, the z axis is

the striction curve, and p = h¢/J ¢, reciprocal of turn rate (Exercise13 of Section 6).

19. Use W = ||xu ¥ xv||.

Section 5.7

1. K = (1 - x2)(1 + x2exp(-x2))-2. Hence K > 0 ¤ -1 < x < 1.3. In a canonical parametrization, if g is constant, the profile curve is

orthogonal to the axis, so the surface M is part of a plane. Otherwise,K = 0 ¤ h≤ = 0 ¤ h¢ is constant. If h¢ = 0, the profile curve lies in aline parallel to the axis, so M is part of a cylinder. If h¢ π 0, the profilecurve is a slanting line, so M is part of a cone.

5. M has parametrization x(r, v) = (rcosv, rsinv, f(r)). Then E = 1 + f ¢2,F = 0, G = r2, and WL = rf ≤, WM = 0, WN = r2f ¢, with W 2 = EG -F 2 = r2(1 + f ¢2).

Â =v p ai i i2 0.

Z x a Ui i i= Â( ) .

7. (a) h(u) = asinh(u/c) satisfies the given differential equation with K = -1/c2. Use the integral formula for g(u). Then as u Æ 0, the slope angle tanj = h¢/g¢ approaches .The curve becomes vertical when g¢ = 0, hence the integrand of gvanishes. There cosh2(u*/c) = c2/a2, so

(b) h(u) = ce-u/c satisfies the differential equation and initial conditionin Example 7.6.

Chapter 6

Section 6.1

1. (a) a ≤ = w12(T)E2 + w 13(T)E3. Hence a ≤ is normal to M if and onlyif w12(T ) = 0.

3. Apply the symmetry equation to E1, E2. Then use Corollary 1.5.

Section 6.2

1. (a) q 1 = dz, q 2 = rdJ.(d) K = 0 and H = -1/2r.

Section 6.3

1. If K = H = 0, then k1k2 = k1 + k2 = 0. Thus k1 = k2 = 0, so S = 0.3. In the proof of Liebmann’s theorem, replace the constancy of K = k1k2

by that of 2H = k1 + k2.5. In the case k1 π k2, use Theorem 2.6 to show that, say, k1 = 0. By Exer-

cise 2 the k1 principal curves are straight lines. Show that the k2 princi-pal curves are circles and that the (k1) straight lines are parallel in R3.

Section 6.4

1. (d) fi (b): If z is an arbitrary tangent vector at p, write z = av + bw.Then

F a F abF F b F

a ab b

* * * • * *

• .

z v v w w

v v w w z

2 2 2 2 2

2 2 2 2 2

2

2

= + +

= + + =

h a u c c amax sin * .= ( ) = -h 2 2

a c a c a c a/ / / /( ) - = -1 2 2 2 2


3. (b) Monotone reparametrization does not affect length of curves.(c) By the definition of r, given any e > 0 there is a curve segment a

from p to q of length < r(p, q) + e, and an analogous b for q andr. Combining a and b gives a piecewise differentiable curve segmentfrom p to r. (If only everywhere-differentiable curves are allowed,there is no change in r, but proofs are harder.)

5. (a) Define F(a (u) + vT(u)) = b(u) + vT(u).(b) Choose b in R2 with plane curvature equal to k.

7. By the exercise mentioned, a shortest curve in R2 joining the pointsparametrizes a straight line segment. Thus any curve in M joining thepoints has length L > 2.

9. F*((F-1)*v) = (FF-1)*v = I*v = v. Since F is an isometry, ||(F-1)*v|| = ||v||.11. Write F(x(u, v)) = (f(u), g(v)) for suitable parametrizations.13. For y, show that the conditions E = G and F = 0 are equivalent

to g¢ = cosg, which has solution g(v) = 2tan-1 (ev) - p /2 such that g(0) = 0. Use criteria suggested by Exercise 8.

15. F(x(u, v)) = ( f(u)cosv, f(u) sinv), where x is a canonical parametriza-

tion and

Section 6.5

1. First show that a is a geodesic if and only if w12(a ¢) = 0. Let 1, 2

be the transferred frame field, with connection form 12. Since 1 =F*(a¢) = F(a)¢, Lemma 5.3 gives

3. There is no local isometry of the saddle surface M (-1 � K < 0) onto acatenoid with -1 � < 0—or vice versa—since K has an isolatedminimum point, at 0, while takes on each of its values on entirecircles. Many other examples are possible.

5. (b) Follows from Lemma 4.5, since computation for xt shows Et =cosh2 u = Gt and Ft = 0.

(d) For Mt, Ut = (s, -c, S)/C, so the Euclidean coordinates of Ut are independent of t.

7. A local isometry must carry minimum points of KH to minimum pointsof KC, and also preserve orthogonality and geodesics.

Section 6.6

1. (b) q q w12

2 122 2 2

1 1 1 1= + = = + = +( )u du u dv dv u K u, , ,/ / .

KK

0 12 12 12 12= ¢( ) = ( ) ¢( ) = ¢( )( ) = ( )¢w a w a w a w aF F F* * .( )

EwEE

f u dt h tu

( ) = ( )ÊË

ˆ¯Úexp .

1

x


3. (b) Substitution into dw 13 = w 12 Ÿ w 23 leads to

Section 6.7

1. 1 + fu2 + fv

2 � 1.

3. (a)

(b) xu ¥ xv points inward, and thus

. Hence

5. F carries positively oriented pavings of M to positively oriented pavingsof N. Apply the suggested exercise to each 2-segment.

Section 6.8

1. (a) F*(du Ÿ dv) = F*(du) Ÿ F*(dv) = df Ÿ dg = (fudu + fvdv) Ÿ (gudu +gvdv) = (fugv - fvgu)du Ÿ dv.

(b)3. (a) Recall that G* ª -S. Let e1, e2 be a principal frame at a point of M.

Then G*(e1) • G*(e2) = 0. Thus G is conformal if and only if||G*(e1)||2 = ||G*(e2)||2 > 0 at every point.

5. Using a canonical parametrization,

7. (a) For a small patchlike 2-segment,

If this always equals , then taking limits as x shrinks to a

point p gives JF (p) = ±1. F must be one-to-one, for otherwise two smallregions of total area 2e could map to a single region of area e.

Conversely, we can suppose F is orientation-preserving; hence JF = 1.Then use Exercise 5 of Section 7.

(b) An isometry carries frames to frames. We have seen that cylindricalprojection of a sphere is area-preserving (Exercise 6 of Section 7).

A dMx

x( ) = ÚÚ

A F dN J dMF

Fxx x

( )( ) = = ±( )ÚÚ ÚÚ .

KdM dv h h h ds

h b h a

a

b

a b

ÚÚ Ú Ú= - ¢¢( )

= - ¢( ) - ¢( )( )= -( )

0

2

1 1

1

1

2

2

p

p

p j jsin sin .

x x x* .dM dM du dv EG F du dvu v( ) = ( ) Ÿ = ± - Ÿ, 2

v T RrTÚ = - ( ) = -area 4 2p .EG F= - - 2

U u v u v• x x x x¥ = - ¥v dv R r Rr u du R r

TÚÚ Ú Ú= + +( ) = +( )0

22 2 2 2 2

0

2

2 4p p

pcos .

LL N

vv

v

EE G

HE= +ÊË

ˆ¯ =

2.


9. (a) See text.(b) See Example 4.3(1) of Chapter 5.(c) First show that on one of the vertical lines, exactly four directions

are omitted by U. Total curvatures: -4p, -•, -•.

13.

Section 6.9

5. (a)

.

7. (a) Example 4.3(2) of Chapter 5 shows that K has a unique minimumat 0. Hence every Euclidean symmetry F must carry 0 to 0, so F isan orthogonal transformation C.

(b) C must carry asymptotic unit vectors to asymptotic unit vectors,

and carry Uz to ±Uz. One such C is .

Chapter 7

Section 7.1

1. (a) The speed squared is ·a ¢, a ¢Ò = a ¢ • a ¢/h2(a).(b) ·hUi, hUjÒ = Ui • Uj = d ij.

3. (a) The definition J(e1) = e2, J(e2) = -e1 is independent of the choiceof positively oriented frame field e1, e2, since for another positivelyoriented frame field,

and this implies J(ê1) = ê2, J(ê2) = -ê1. Then for v π 0, choose e2 sothat e1 = v/|v|, e2 is positively oriented.(b) V = f1E1 + f2E2, with f1, f2 differentiable. For the other two rela-

tions, first replace arbitrary vectors by e1, e2.(c) If E1, E2 is positively oriented for dM, then E1, -E2 is positively ori-

ented for -dM.

ˆ cos sin ˆ sin cose e e e e e1 1 2 2 1 2= + = - +J J J J, ,

0 1 0

1 0 0

0 0 1

-

-

Ê

Ë

ÁÁÁ

ˆ

¯

˜˜˜

F = =

-

Ê

Ë

ÁÁÁ

ˆ

¯

˜˜˜

C

1 0 0

0 0 1

0 1 0

TC K r W r dr= ( ) ( ) = -•

Ú2 40

p p.


5. (a) Expand ||v ± w||2 = ·v ± w, v ± wÒ.(b) Compute ·v, wÒ with the vectors expressed in terms of xu and xv.(c) Direct computation with a ¢ = a¢1xu + a¢2xv, yields the same result as

applying ds2 to a ¢, since du(a ¢) = a¢1, dv(a ¢) = a¢2.7. We have F*(U1) = fuU1 + guU2, F*(U2) = fvU1 + gvU2. If F is confor-

mal and orientation-preserving, then using Exercise 6,

So the Cauchy-Riemann equations hold. Conversely, if the Cauchy-Riemann equations hold, then

This proves F is conformal (and shows that |dF/dz| is the scale factor).F is orientation preserving since JF = fugv - fvgu = f 2

u + g2u > 0.

9. (F*(v) • F*(w))/h2F(p) = v • w/h2(p).

Section 7.2

3. A = pa2/(1 - a2/4); hence total area is infinite.5. Since x is an isometry, the area of T0 is the same as the area of a Euclid-

ean rectangle with sides 2pR and 2pr. Hence A(T0) = 4p2Rr, the sameas A(T ).

7. (c) Evidently, i = cq i, and hence 12 = w 12 follows by uniqueness inthe first structural equations.

(d) d = 1 Ÿ 2 = c2q1 Ÿ q2 = c2dM.(e) Theorem 2.1 defines K.

9. (b) Since qi = qi (xu)du + qi (xv)dv = ·Ei, xuÒdu + ·Ei, xvÒdv, we find

(c) Substitute w12 = Pdu + Qdv and preceding results into the firststructural equations.

(d) Substitute into the second structural equation.

q q1 2= + =E du F E dv W E dv/ , / .

qqM

wq

F U F U f f g g f f f f

F U F U f g dF dz f g F U F U

u v u v u v u v

u u v v

* * ,

* , * * * .

1 2

1 12 2 2 2 2

2 2

0( ) ( ) = + = - =

( ) ( ) = + = = + = ( ) ( )

, and

,

f U g U F U

F JU

J F U

J f U g U

g U f U

v v

u u

u u

1 2 2

1

1

1 2

1 2

+ = ( )= ( )= ( )= +( )= - +

*

*

*

.


11. (b) K = -2/cosh3(2u).13. (a) To define tensorK, first simplify the square root of E(u, v)G(u, v)

- F(u, v)2 to get W(u, v).(b) The formulas for E, F, G in the Appendix are valid for arbitrary

n, so evaluate tensorK on the functions ee[x],ff[x],gg[x]for Mathematica; ee(x),ff(x),gg(x) for Maple.

Section 7.3

1. (a) First find the dual 1-forms.(b) a≤ = -cot ta ¢.(c) b¢ = c/(st)E1 + 1/tE2, and ·b¢, b¢Ò¢ = -2/(s2t3).

3. From the proof of Lemma 3.8,5. (a) Let w12 be the connection form of a frame field on D. Since

Stokes’ theorem gives . From

the text,

7. (a) If W = fE1, then —V(W ) = V [ f ]E1 + fw12(V )E2, hence

On the other hand,

But [ f(F-1)] = (F*V)[ f(F -1)] = V[ f(F -1F )] = V[ f ], and

where the last (crucial) step uses Lemma 5.3 of Chapter 6. This com-pletes the proof.

Section 7.4

1. Since a ≤ = 0, we get which is 0 if and only if h≤ = 0.3. If L is a line in the xy plane, consider the Euclidean plane passing through

both L and the north pole n of S 0; then use stereographic projection.5. (a) Use Exercise 5 of Section 3. Since a ¢ is parallel on a, –a (a ¢(a),

a ¢(b)) is the holonomy angle ya.(b) (ii) The image of the Gauss map of a paraboloid is an open

hemisphere of S, hence any (finite) simple region in it has total curvature <2p.

7. (a) Fix p0 Œ M, and let U consist of all points that can be joined to p0

by a broken geodesic—include p0 in U. If p Œ U, then by the givenfact, U contains an e-neighborhood of p. Thus U is open. In asimilar way, M - U is open. Since U is not empty, M = U.

a ah h h( )¢¢ = ¢( ) ¢¢,

w w w w12 12 12 12V F V F V V( ) = ( )( ) = ( )( ) = ( )* * .

V

— ( ) = — ( )( ) = ( )[ ] + ( ) ( )- - -V VW f F E V f F E f F V E1

11

11

12 2w .

— = — ( )( ) = [ ] + ( ) ( )-v VW F W V f E f F V E* .1

112 2w

w ya

a12Ú = - .

wa

12Ú ÚÚ= - K dMg

d K dMw12 = - ,

w12 3 3 1 3 1 3Y E E E E S Y E Ey( ) = -— ◊ = ( ) ◊


Section 7.5

1. The coordinates u, v have E = G = 1/v2, F = 0, hence are Clairaut. Withthe suggested reversals, geodesics are given by

Set w = 1 - c2v2 and integrate to get . Consequently,(u - u0)2 + v2 = 1/c2.

3. At the meeting point, u1 = a1(t1). Since , the conditionG(u) = c2 implies sinj = ±1. Thus a1¢(t1) is tangent to the barrier curve,so a1¢(t1) = 0.

The geodesic equation A1 = 0 in Theorem 4.2 reduces to a1≤ =Gu a2¢2/(2E ). At the meeting point, Gu π 0 since barriers are not geodesic, and a2¢ π 0 since a1¢ = 0. Thus a2≤(t1) π 0. This means that a leaves the barrier curve instantly, remaining on the same side ofit.

5. (a) By Exercise 4, tangency to the top circle implies slant c = R (largerof the radii of T ). Except for the inner and outer equators, no par-allel is geodesic. Hence a leaves the top circle, necessarily enteringthe outer half of T. As h increases, sinj decreases; hence a meetsand crosses the outer equator. By symmetry, it returns to tangencywith the top circle.

(b) Crossing the inner equator implies slant c < R - r.7. Evidently all meridians approach the rim on a finite parameter interval.

In view of the exercises above, so do all other geodesics; even if initiallymoving away from the rim, they will be turned back by a barrier curve.They cannot asymptotically approach a parallel, since no parallels aregeodesic.

9. (a) E(u) = ee(u), G(u) = gg(u) will be given (for abstract surfaces) orcomputed (for surfaces in R3).(Mathematica):clair[u0_,v0_,c_,tmin_,tmax_]:=NDSolve[{u’[t]==Sqrt[gg[u[t]]–c^2]/Sqrt[ee[u[t]]*gg[u[t]]],v’[t]==c/gg[u[t]],u[0]==u0,v[0]==v0},{u,v},{t,tmin,tmax}]

(b) ParametricPlot3D[Evaluate[x[u[t],v[t]]/.nsol],{t,tmin,tmax}]

where nsol is an explicit return from clair. (Delete “3D” in theabstract case.)

c G a= ( )1 sinj

u u w c- =0 m /

dudv

c G

E E c

cv

c v=

±-

=±-2 2 21

.


(Maple)(a) clair:=(u0,v0,c)–>dsolve({diff(u(t),t)=

(gg(u(t))–c^2)^(1/2)/(ee(u(t))*gg(u(t))^(1/2),diff(v(t),t)=c/gg(u(t)),u(0)=u0,v(0)=v0},{u(t),v(t)},type=numeric).

(b) With plots installed, if nsol is an explicit return from clair,odeplot(nsol,x(u(t),v(t)),tmin..tmax)

11. (b) Since G(0) = f(0)2 = (3/4)2, the slant of this geodesic is ±3/4.13. Since a ¢ = a ¢1xu + a¢2xv, we have . Hence

cos2j = (U(a1) + V(a2))a ¢12, and sin2 j is similar. Thus we must show thatthe function f = (U(a1) + V(a2)(U(a1)a¢22 - V(a2)a¢12) is constant.Compute f ¢. The geodesic equations from Theorem 4.2 then give f ¢ = 0.

Section 7.6

1. In (a) and (c) the surface is diffeomorphic to a sphere, so TC = 4p. In(b), there are four handles, so TC = -12p.

3. If h = 0, then M is a sphere, so TC > 0. If h = 1, then M is diffeo-morphic to a torus; hence TC = 0. If h � 2, then TC < 0.

5. (c) For each polygon, draw lines from a central point to each vertex.Thus each original n-sided face is replaced by n faces, and there aren new edges and one new vertex. Thus for each polygon, the effecton c(M ) is 1 Æ 1 - n + n, so there is no change.

7. The area of x(R) is . Three of the four edges are geodesics.9. We count e = 6f /2 = 3f and v = 6f /3 = 2f; hence c = 0. So this is

impossible on the sphere, but a suitable diagram shows that the torushas such a decomposition.

Section 7.7

1. Follows from Theorem 7.5 since a polygon has Euler characteristic +1.3. (a) By the Gauss-Bonnet theorem, M is diffeomorphic to a sphere; hence

if two simply closed geodesics do not meet, they bound a region.5. (a) The angle function from any X to Vt depends continuously on t;

hence the index depends continuously on t. But a continuousinteger-valued function on an interval is constant.

(b) Use (a).7. (a) Approximate closely by a genuine polygon. In the limit, the in-

terior angles will all be p. Hence by Exercise 1, -An/r2 = (2 - n)p,so An = (n - 2)pr2.

pr 2 4 2/ ( )

cos /j a= ¢ = ¢, xu E E a1


(b) As n Æ •, An Æ •, so H(r) has infinite area.9. (a) Let h = ||Va || > 0. Then f = hcosj, g = hsinj, so the integrand

reduces to j¢.11. (a) The equations u¢ = -u, v¢ = v have general solutions u = Ae-t, v =

Bet, so A = a, B = b.(b) Since uv = ab, the integral curves parametrize hyperbolas (when

ab π 0); this is a meeting of two streams, with index -1.(c) For the circle a(t) = (cos t, sin t), the integrand reduces to -1.

13. (a) (Mathematica): numsol[u0_,v0_,tmin_,tmax_]:=NDSolve[{u’[t]==2u[t]^2–v[t]^2,v’[t]==–3u[t]*v[t],u[0]==u0,v[0]==v0},{u,v},{t,tmin,tmax}]draw[u0_,v0_,tmin_,tmax_]:=ParametricPlot[Evaluate[{u[t],v[t]}/.numsol[u0,v0,tmin,tmax]],{t,tmin,tmax}]

(b) (Maple): Take X = (1, 0); hence J(X) = (0, 1). Now apply Exercise9. Evaluation on the circle a(t) = (cos t, sin t) givesf:=t–>2*cos(t)^2–sin(t)^2,g:=t–>–3*cos(t)*sin(t)The integrand iswint:=t–>(f(t)*diff(g(t),t)–g(t)*diff(f(t),t))/(f(t)^2+g(t)^2) and int(wint(t),t = 0..2*Pi) is -4p, so the index is -2.

Chapter 8

Section 8.1

1. (a) If q is in a normal e-neighborhood N of p, then by Theorem 1.8,the radial geodesic from p to q has length r(p, q) < e. If q is not inN, then any curve from p to q meets every polar circle of N; hencer(p, q) � e.

3. n(x, y) = (rcos(x/r), rsin(x/r), y). To get the largest normal e-neighbor-hood, fold an open Euclidean disk of radius pr around the cylinder.

5. (a) Any geodesic starting at p is initially tangent to a meridian;hence (by the uniqueness of geodesics) parametrizes that meridian. It follows that the entire surface is a normal neighbor-hood of p.

7. (a) By the triangle inequality, r(p, q) > r(p0, q) - r(p0, p). Reversing pand q, we conclude that r(p, q) > |r(p0, q) - r(p0, p)|.


(b) Show that if r(p0, p) < e and r(q0, q) < e, then it follows that |r(p0, q0) - r(p, q)| < 2e.

Section 8.2

1. Let M be an open disk in R2.3. We can assume that C is parametrized by a(u) + vU3, with a a unit-

speed curve. If a is (smoothly) closed, let s have the same arc lengthand parametrize a circle in R2. Then a(u) + vU3 Æ s(u) + vU3 is anisometry. Circular cylinders of different radii are not isometric sincetheir closed geodesics have different lengths.

If a is one-to-one, then since it is a geodesic of C it is defined on theentire real line. Then a(u) + vU3 Æ (u, v) is an isometry onto R2.

5. The profile curves all approach either a singularity of the curve or theaxis of rotation. Only for the sphere was the axis met orthogonally, thusgiving S as an augmented surface of revolution.

Section 8.3

1. For k = -1/r2, the general solution of the Jacobi equation g≤ - g/r2 =0 can be written as g(u) = Acoshu/r + Bsinhu/r. The initial conditionsthen determine A and B.

3. (a) L(e) = 2psinhe.5. (a) xu(0, v) = X(v), and since x(0, v) = gx(v)(0) = b(v), we have xv(0, v) =

b¢(v). Thus EG - F2 is nonzero when u = 0, hence also for |u| small.(b) (iii) b as base curve, X = d.

7. (a) The u-parameter curves of x are meridians of longitude.(b) Since K = 0, the Jacobi equation becomes . Hence is

linear in u, and it follows that .

Section 8.4

1. Let E be the due-east unit vector field on the sphere S (undefined at thepoles). If A is the antipodal map, then A*(E) = E, so E transfers to Pvia the projection S Æ P. The unique singularity has index 1.

3. The condition implies F(M) = N. If q is in N, then each point of F-1(q)has a neighborhood mapped diffeomorphically onto a neighborhood ofq. The intersection V of all these neighborhoods of q is evenly covered;the condition prevents its lifts from meeting.

G u v v ug,( ) = - ( )1 k

GG uu ( ) = 0


5. (a) Since covering maps are local diffeomorphisms and T is orientable,T cannot be covered by a nonorientable surface (Exercise 3 ofSection 4.7). Thus any compact connected covering surface M of Tmust also have c(M ) = 0. Hence by Theorem 6.8 of Chapter 7, Mis a torus.

(b) For the usual parametrization of T, let F(x(u, v)) = x(nu, v).

Section 8.5

1. If F: M Æ N is an isometry, define f: I(M) Æ I(N) by f(G) = FGF -1.Show that f is a homomorphism and is one-to-one and onto.

3. Suppose p π q in M. Then any geodesic segment s from p to q hasnonzero speed, so F(s) is a nonconstant geodesic of N. If F(p) = F(q),there are two geodesics from this point to the midpoint of F(s).

5. (a) Given points p and q in M, if F and G are isometries such that F(p0) = p and G(p0) = q, then the isometry GF-1 carries p to q.

(c) Given frames e1, e2 at p and f1, f2 at q, let F and G be isometries suchthat F(p) = p0 and G(q) = p0. By hypothesis, there is an isometry H that carries the frame F*(e1), F*(e2) to G*(f1), G*(f2). Then theisometry G-1HF carries e1, e2 to f1, f2.

7. (a) Since C on R3 is linear, C(-p) = -C(p). Then the mapping {p, -p} Æ {C(p), -C(p)} has the required properties.

(b) Because F is a local isometry, any two frames on P can be writtenas F*(e) and F*(f), where e and f are frames on S. By Exercise 6there is an orthogonal transformation C of R3 such that C*(e) = f.Now use FC = CpF.

9. (a) .(b) Use Exercise 1. The only derived isometries are those of the form

F(x(u, v)) = x(±u + a, ±v).11. Calculate ||F(p)||.

Section 8.6

1. (a) One handle implies c(M ) = 0, but by Gauss-Bonnet, K < 0 impliesc < 0.

(b) By the Gauss-Bonnet formula, the angle sum for a k = -1 rectan-gle can never be 2p.

3. Only the projective plane satisfies all three axioms; the others fail onaxiom (ii), and S also fails (i).

5. For D, let e1, e2 be the frame at the common vertex of a and b such that e1 is tangent to a, and cosJe1 + sinJe2 is tangent to b. Let F bethe isometry carrying the frame e1, e2 to the corresponding frame on D¢.

x u v u v v E F G, , , has( ) = +( ) = = =-sinh , ,1 21 1 0 1


Section 8.7

1. In the proof of assertion (3) in Lemma 7.4, the Jacobi equation nowreduces to g≤ = 0, so the initial conditions then give g(u) = u.

3. For a point p0 in M, the functions p Æ r(p0, p) and (when relevant) p Æ d(p0, p) are both continuous, hence take on maximum values.Then use the triangle inequality.


B

Barrier curve, 361 (Ex. 2)Basis formulas, 269Bending, 283, 286

helicoid to catenoid, 293 (Ex. 5)Binormal, 59, 69, 72Bonnet’s theorem, 438–439Boundary

of a 2-segment, 176of a polygonal region, 377

Bounded 198 (Ex. 5), 443, 449 (Ex. 3)Bracket operation, 208 (Ex. 9)Bugle surface, 259–260, 262 (Ex. 8), 299–313

C

Canonical isomorphism, 45, 62Canonical parametrization, 256Cartan, E., 43, 85, 95Cartesian product, 200 (Ex. 15)Catenoid, 254

Gauss map, 308Gaussian curvature, 254, 256local isometry onto, 283as minimal surface, 254–255total Gaussian curvature, 305–306,

309Center of curvature, 67 (Ex. 6), 79

(Ex. 11)Circle, 61, 65Clairaut parametrization, 33Classical geometries, 440

▼

▲Index

495

A

Acceleration of a curvein R3, 54–55, 70in a surface, 203, 341

Adapted frame field, 264Algebraic area, 307–308, 376 (Ex. 8)All-umbilic surface, 275–276Alternation rule, 28, 48, 159–160Angle, 45, 322

coordinate, 224exterior, 366interior, 366oriented, 311turning, 366

Angle function, 52 (Ex. 12)along a curve, 311–312on a surface, 324slope, 68 (Ex. 6), 351

Antipodal mapping, 173 (Ex. 5), 194Antipodal points, 194Arc length, 52–53, 231 (Ex. 5)

parametrization, 53Area, 297–303Area form, 301, 312Area-preserving mapping, 304 (Ex. 6), 313

(Ex. 7)Associated frame field

of a coordinate patch, 294, 336of a vector field, 312

Asymptotic curve, 243–244Asymptotic direction, 242–243Attitude matrix, 47–48, 91–92

Classification of compact surfaces, 423nonorientable, 422orientable, 371

Closed differential form, 164–165Closed surface in R3, 198 (Ex. 4), 404Codazzi equations, 267, 272Column-vector conventions, 105Compactness, 184Compact surface, 184–185, 276–277, 280Complete surface, 350

geodesics, 400Composite function, 4Cone, 146 (Ex. 3), 233 (Ex. 13)Conformal geometric structure, 323, 331Conformal mapping, 286, 288 (Ex. 8)Conformal patch, 288 (Ex. 8)Congruence of curves, 121, 127 (Ex. 5)

determined by curvature and torsion,121–123, 126

Congruence of surfaces, 314–315Conjugate point, 405–410Connected surface, 184, 192 (Ex. 9)Connection equations

on Euclidean space, 89, 266on a surface, 267, 38

Connection formson Euclidean space, 89on a surface, 266, 289, 295, 324

Conoid, 251 (Exs. 17, 18)Consistent formula for a mapping, 174

(Ex. 13)Constant curvature surface

flat, 435–437negative, 437–438positive, 435standard, 433

Continuous function, 369, 384Coordinate angle, 224Coordinate expression, 149Coordinate patch, See PatchCoordinate system on a surface, 165 (Ex. 7),

295Covariant derivative

Euclidean, 81–84, 121 (Ex. 5)intrinsic, 337–340, 341on a patch, 202–203relation of Euclidean to intrinsic,

343–344Covariant derivative formula, 93 (Ex. 5), 338Covering map, 416–417

dent multiplicity, 419–420Riemannian, 423–424

Covering manifold (surface), 417Critical point, 28Cross product, 48–50, 111, 113Crosscap, 422Cross-sectional curve, 78 (Exs. 7, 8)Curvature, See also Gaussian curvature;

Geodesic curvature; Mean curvatureof a curve in R2, 68 (Ex. 8)or a curve in R3, 58, 69, 72

Curve, 16, 150closed, 188coordinate functions, 150,one-to-one, 21periodic, 21, 156 (Ex. 2)regular, 21in a surface, 150unparametrized, 21–22

Curve segment, 52–53minimizing, 389shortest, 389

Cylinder, 146 (Ex. 4)geodesics, 246

Cylindrical frame field, 85connection forms, 92–93dual 1-forms, 97 (Ex. 3)

Cylindrical helix, 75–76, 78

D

Darboux, G., 85–86Darboux frame field, 248 (Ex. 7)Degree of a form, 29Degree of a mapping, 376 (Ex. 8)Diffeomorphic surfaces, 169, 371Diffeomorphism

of Euclidean space, 40of surfaces, 169,

Differentiability, 4 10 34 150–151Differential form,

closed, 164–165exact, 164–165on a surface, 158–163on R2, 163on R3, 22–23, 28–29pullback of, 171–172

Differential of a function, 25Dini’s surface, 262Direction, 209

496 Index

Directional derivative, 11–12, 155computation of, 12, 26

Diskpolar, 414 (Ex. 2)smooth, 192 (Ex. 6)

Distribution parameter, 249 (Ex. 11)Domain, 1Dot product, 43–44, 85, 224

preserved by isometries, 116Dual 1-forms, 94–95, 266, 289, 297, 324Dupin curves, 222 (Ex. 5)

E

E, F, G (metric components), 146 (Ex. 2),224, 234 (Ex. 18), 337 (Ex. 4)

Edge (curve), 177, 369Efimov’s theorem, 439Ellipsoid, 148 (Ex. 9)

Euclidean symmetries, 319 (Ex. 8), 427Gaussian curvature, 236–238isometry group, 427umbilics on, 240 (Ex. 7)

Elliptic paraboloid, 149 (Ex. 10), 232 (Ex. 6)

geodesics, 356–357Ennepers surface, 250–251 (Exs. 15, 16),

313–314 (Exs. 10, 11)e-neighborhood, 44Euclid, 360Euclidean coordinate functions, 9, 16, 24, 33,

55Euclidean distance, 44, 50 (Ex. 2), 383

(Ex. 1), 399 (Ex. 7)Euclidean geometry, 116–117Euclidean plane, 5Euclidean space, 3–5

natural coordinate functions, 4natural frame field, 9

Euclidean symmetry group, 319 (Ex. 6)Euclidean vector field, 153, 158 (Ex. 12)Euler characteristic, 370–371Euler’s formula, 214Evolute, 79 (Exs. 17, 18)Exact differential form, 164–165Exponential map, 389–390

of the real line, 417Exterior angle, 366Exterior derivative, 30, 33 (Ex. 7),

161–163

F

Faces, 369Fary-Milnor theorem, 81Fenchel’s theorem, 80 (Ex. 18), 309Flat surface, 220Flat torus, 330

imbedded in R4, 430Focal point, 416–417 (Ex. 6)Form, See Differential formFrame, 45Frame field

adapted, 264on a curve, 126on R3, 85natural, 9principal, 271on a surface, 264, 389transferred, 290–291

Frame-homogeneous surface, 428, 440Frenet, F., 84Frenet apparatus, 66 (Ex. 1)

preserved by isometries, 118for a regular curve, 69for a unit speed curve, 58–59

Frenet approximation, 63, 68 (Ex. 9)Frenet formulas, 60, 69, 350Frenet frame field, 59Function, 1–2

one-to-one, 2onto, 2

Fundamental form, 222 (Ex. 4), 329 (Ex. 5)

G

Gauss, K. F., 263Gauss-Bonnet formula, 367–368Gauss-Bonnet theorem, 372–375, 378Gauss equation, 267Gauss map, 207–208 (Exs. 4–8), 308–309Gaussian curvature, 216–218, 329, See also

Specific surfacesformulas for

direct, 216–217, 219–220, 226, 273–274,296–297, 336–337 (Ex. 9)

indirect, 219, 269–270, 413, 414 (Ex. 2)and Gauss map, 308and holonomy, 345 (Ex. 5)of an implicitly defined surface, 236–237isometric invariance, 291–292

Index 497

Gaussian curvature (continued)and principal curvatures, 216sign, 216–218

Geodesic curvature, 248 (Ex. 7), 350Geodesic lift property, 423Geodesic polar mapping, 391Geodesic polar parametrization, 391–393Geodesics, 245–246, 346, See also Specific

surfacesbroken, 353 (Ex. 7)closed, 246coordinate formulas for, 351–352existence and uniqueness, 348–349locally minimizing, 405maximal, 349minimizing, 394, 397–398periodic, 246preserved by (local) isometrics, 293

(Ex. 1), 425Geographical patch, 140Geometric surface, 322. See also Constant

curvature surfacesinextendible, 403

Gradient, 33 (Ex. 8), 51 (Ex. 11) as normalvector field, 153

Group, 106Euclidean, 106 (Ex. 7)Euclidean symmetry, 319–320isometry, 427orthogonal, 106 (Ex. 8)

H

Hadamard’s theorem, 448Handle, 371Hausdorff axiom, 192 (Ex. 10), 193, 193

(Ex. 11)Helicoid, 146 (Ex. 5)

local isometries, 284–286, 294 (Ex. 7)patch computations, 227–228

Helix, 16, 60–61, 119–120, 124Hilbert’s lemma, 278Hilbert’s nonimbedding theorem, 439Holonomy, 343

angle, 343Homeomorphic surfaces, 369Homogeneous surface, 428Homotopy, 188

free, 191Hopf’s degree theorem, 379 (Ex. 8)

Hopf-Rinow theorem, 400Hyperbolic paraboloid, 149 (Ex. 10), 232

(Ex. 6)Hyperbolic plane, 332–333, 335 (Ex. 4)

completeness, 397geodesics, 358–359frame-homogeneity, 364 (Ex. 14), 440

Hyperboloids, 148–149 (Ex. 9), 232 (Ex. 6),238 (Ex. 1)

I

Identity map, 102Image, 1Image curve, 36Imbedding, 201 (Ex. 16)

isometric, 429Immersed surface, 201 (Ex. 17)Immersion, 201 (Ex. 17)

isometric, 429Improper integral, 303Index of a singularity, See SingularityInitial velocity, 22 (Ex. 6)Inner product, 43, 321–322Integral curve, 200 (Exs. 13, 14)Integral of a function on a surface, 303Integration of differential forms,1-forms over 1-segments, 174–176,

178–180of 2-forms over 2-segments, 177of 2-forms over oriented regions, 303, 303

(Ex. 4)Interior angle, 366Intrinsic distance, 281, 287 (Ex. 3), 387

(Ex. 7)Intrinsic geometry, 289Inverse function, 2Inverse function theorem, 40, 169Isometric imbedding, 429Isometric immersion, 429Isometric invariant, 289, 321Isometric surfaces, 283Isometry of Euclidean space, 100

decomposition theorem, 105determined by frames, 109–110tangent map, 107–108

Isometry group, See also Euclideansymmetry group

of Euclidean space, 107 (Ex. 7)of a geometric surface, 426

498 Index

Isometry of surfaces, 282and Euclidean isometries, 314–316

Isothermal coordinates, 297 (Ex. 2)

J

J (rotation operator), 79 (Ex. 12), 311, 327(Ex. 3)

Jacobi equation, 409–410Jacobian (determinant), 156 (Ex. 3), 161,

306, 312 (Ex. 1)Jacobian matrix, 40Jacobi’s theorem, 407Jordan curve theorem, 352

K

Klein bottle, 436Kronecker delta, 25

L

L, M, N, 228, 230, 234 (Ex. 18)Lagrange identity, 222 (Ex. 6)Law of cosines, 441 (Ex. 4)Leibnizian property, 14Length

of a curve segment, See Arc lengthof a vector, 322. See also Norm

Liebmann’s theorem, 280Line of curvature, See Principal curveLine-element, 328 (Ex. 5)Liouville parametrization, 364

(Ex. 13)Liouville’s formula, 403Local diffeomorphism, 173 (Ex. 6)Local isometry, 283–284, 426

criteria for, 284determined by differential map, 426

Local minimization of arc length, 405–407,408

Loop, 188–189Loxodrome, 234 (Ex. 16)

M

Manifold, 196, 201, 326Mapping of Euclidean spaces, 34–35

of surfaces, 166–167Massey, W. S., 404

Mean curvature, 216, 217–218, 221 (Ex. 3),226, 237, 269–270

Mercator projection, 286 (Ex. 13)Metric tensor, 322

components of, See E, F, Gcoordinate description, 324, 328 (Ex. 4)induced, 323

Milnor, T. K., 439Minding’s theorem, 416 (Ex. 8)Minimal surface, 221

examples, 255. See also Enneper’s surface;Scherk’s surface

Gauss map, 313 (Ex. 9)ruled, 251–252 (Ex. 19)as surface of revolution, 255

Minimization of arc length, 389local, 405

Möbius band, 1871, 198–199 (Exs. 8–10)complete and flat, 436

Monge patch, 133, 229 (Exs. 2, 3)Monkey saddle, 137, 218, 314 (Ex. 13)

Gaussian curvature, 230 (Ex. 7)

N

Natural coordinate functions, 4Natural frame field, 9Neighborhood, 44, 131. See also Open set

normal, 390Norm, 44, 45, 85Normal coordinates, 398 (Ex. 2)Normal curvature, 209–212, 232 (Ex. 11)Normal plane, 69 (Ex. 9)Normal section, 210–211Normal vector field, 153–154

O

One-to-one, 2Onto, 2Open interval, 16–17Open set

in Euclidean space, 5, 44in a surface, 158, 192

Orientable surface, 186–187, 198 (Ex. 7),301

Orientationdetermined by an area form, 301determined by a unit normal, 186, 203,

208

Index 499

Orientation (continued)of frames, 110, 311of a patch, 301–303of a paving, 302–303of a region, 208

Orientation of a tangent frame fieldsopposite, 324–3025same, 324–325

Orientation covering surface, 198 (Exs. 6, 7),423

Orientation-preserving (-reversing) isometry, 112, 115, 116, (Ex. 6),306

Orientation-preserving (-reversing)reparametrization of a curve, 54,180–181

monotone, 57 (Ex. 7)Oriented angle, 311–312Oriented boundary, 177–178, 377Orthogonal coordinates, 295–296

Gaussian curvature formula in, 296Orthogonal matrix, 48Orthogonal transformation, 102, 104Orthogonal vectors, 45, 65–66, 322Orthonormal expansion, 47Orthonormal frame, See FrameOsculating circle, 67 (Ex. 6)Osculating plane, 63, 68 (Ex. 9)Osserman, R., 314

P

Paraboloid, See Elliptic paraboloidParallel curves, 58 (Ex. 10), 69 (Ex. 11)Parallel postulate, 360–361Parallel surfaces, 223 (Ex. 7)Parallel translation, 342Parallel vector field, 56, 341–342Parallel vectors in Euclidean space, 6Parameter curves, 143Parametrization

of a curve, 22of a surface, 142–143decomposable into patches, 173 (Ex. 7)

Partial velocities, 140, 141, 153, 168–169Patch, 130

abstract, 193geometric computations in, 224–226Monge, 133orthogonal, 232 (Ex. 8), 294

principal, 233 (Ex. 8), 293 (Ex. 3)proper, 131, 136, 158 (Ex. 14)

Patchlike 2-segment, 297Paving, 300, 302–303Planar point, 218Plane curvature, 68 (Ex. 8), 79 (Ex. 12)Plane curve, 63

Frenet apparatus, 68 (Ex. 8)Plane in R3, 62, 137 (Ex. 2), 245, 274–275

identified with R2, 132Poincaré, H., 376Poincaré half-plane, 327 (Ex. 2), 399

(Ex. 8)geodesics, 361 (Ex. 1)isometric to hyperbolic plane, 362–363

(Ex. 8)Poincaré-Hopf theorem, 381–382, 422Poincaré’s lemma, 189Point of application, 6Pointwise principle, 9Polar circle, 392Polarization, 103–104, 328 (Ex. 5)Polygonal decomposition, 370Polygonal region, 376

boundary segment, 376Pregeodesic, 352Principal curvatures, 212

as eigenvalues, 213formula for, 214, 220

Principal curve, 240–241, 247–249Principal direction, 212Principal frame field, 271–272Principal normal, 59, 69, 72, 350Principal vectors, 212, 232 (Ex. 9)

as eigenvectors, 213Projective plane, 194–195, 334

frame-homogeneity, 432 (Ex. 7)geodesics, 352–353 (Ex. 6)isometric imbedding of, 432 (Ex. 10)topological properties, 197 (Ex. 2)

Pseudosphere, See Bugle surfacePullback

of a form, 170–171of a metric, 323

Push forward of a metric, 333

Q

Quadratic approximation, 214–215Quadric surface, 148

500 Index

R

Rectangular decomposition, 369Rectifying plane, 68 (Ex. 9)Reflection, 113Regular curve, 22Regular mapping, 39, 169Reparametrization of a curve, 19–20

monotone, 57 (Ex. 7)orientation-preserving, 54orientation-reversing, 54unit-speed, 53

Riemann, B., 321, 360Riemannian geometry, 326Riemannian manifold, 326Rigid motion, See Isometry of Euclidean

spaceRigidity, 318 (Ex. 1)Rotation, 101, 113, 115 (Ex. 4), 116 (Ex. 6)Ruled surface, 145, 233 (Ex. 12), 244, 313

(Ex. 8)flat, 233 (Ex. 13)noncylindrical, 249–250 (Exs. 11–13)

Ruler function, 323Ruling, 145

S

Saddle surface, 147 (Ex. 6)Euclidean symmetries, 319 (Ex. 7)Gauss map, 314 (Ex. 12)patch computations, 229–230

Scalar multiplication, 8, 9Scale change, 336 (Ex. 7)Scale factor, 286Scherk’s surface, 239 (Ex. 5)

Gauss map, 313 (Ex. 9)Schwarz inequality, 45, 322Serret, J. A., 84Shape operator, 203–204

characteristic polynomial, 221 (Ex. 4)and covariant derivative, 344and Gauss map, 308–309and Gaussian and mean curvature, 216and normal curvature, 209preserved by Euclidean isometries,

314–315and principal curvatures and vectors, 213proof of symmetry, 226–227, 269 (Ex. 3)in terms of a frame field, 266

Shortest curve segment, 389

Sign of an isometry, 109Simply connected surface, 188Singularity, 380, 385 (Ex. 10)

index, 380, 385 (Ex. 9)isolated, 380removable, 384sources and sinks, 380, 381

Slant of a geodesic, 354Smooth disk, 192 (Ex. 5)Smooth function, 4Smooth overlap, 151–152, 194Speed of a curve, 52Sphere, 133

conjugate points, 407, 410Euclidean symmetries, 319 (Ex. 4), 432

(Ex. 6)Euler characteristic, 370–371frame-homogeneity, 432 (Ex. 6), 440Gaussian curvature, 221, 231 (Ex. 1),

270geodesics, 245, 396–397geographical patch, 140computations, 230 (Ex. 1), 296geometric characterizations, 275, 276, 280holonomy, 343isometries, 319 (Ex. 4)rigidity, 318 (Ex. 1)shape operator, 202–203topological properties, 184, 188, 189

Sphere with handles, 371Spherical curve, 66, 68 (Ex. 10), 81(Ex. 20)Spherical frame field, 87, 97

adapted to sphere, 267–268dual and connection forms, 97

Spherical imageof a curve, 74–75of a surface, See Gauss map

Standard constant curvature surface, 433,435

Stereographic plane, 331Stereographic projection, 167, 169–170, 173

(Ex. 12)as conformal mapping, 288 (Ex. 14)

Stereographic sphere, 331–332Stokes’ theorem, 178–179, 183 (Ex. 13), 377Straight line, 16, 58 (Ex. 11)

length-minimizing properties, 58 (Ex. 11)Structural equations on R3, 95–96

on a surface, 267, 270, 329, 330Support function, 238

Index 501

Surfaceabstract, 195–196geometric, 322immersed, 201 (Ex. 16)in Rn, 335in R3, 131, 429

implicit definition, 133–134simple, 133, 172 (Ex. 3)

Surface of revolution, 135, 243–253area, 303 (Ex. 2)augmented, 138 (Ex. 12)of constant curvature, 257–259, 261–262

(Ex. 7)diffeomorphism types, 191 (Ex. 5)Gaussian curvature, 253, 256geodesics, 3622 (Ex. 4)local characterization, 288 (Ex. 12)meridians and parallels, 135–136parametrization canonical, 256

special, 148 (Ex. 8)usual, 143–144

patch computations, 239–240principal curvatures, 253principal curves, 242, 239–240profile curve, 135total Gaussian curvature, 312–313 (Exs. 5,

6)twisted, 262

Symmetry equation, 267

T

Tangent bundle, 196–197Tangent direction, 209–210Tangent line, 23 (Ex. 9), 63Tangent map, 37, 40 (Ex. 9)

of a Euclidean isometry, 107–109of a mapping of surfaces 168–169, 173

(Exs. 9, 10), 426of a patch, 156 (Ex. 4)

Tangent plane, 152–153Euclidean, 157 (Ex. 9)

Tangent space, 7Tangent surface, 233 (Ex. 13), 335

local isometries, 287 (Ex. 5)Tangent vector,

to R3, 6, 15,to a surface, 152

Theorema egregium, 291–293, 3293-curve, 17, 73–74

Topological invariants (properties), 184–191,370

Torsion of a curve, 60, 69, 72formula for,72sign, 119

Torus of revolution, 144. See also Flat torusEuler characteristic, 370Gauss map, 308Gaussian curvature, 217–218, 254patch computations, 254–255total Gaussian curvature, 305, 309–310usual parametrization, 144–145

Total curvature of a curve in R3, 80 (Ex. 17)Total Gaussian curvature, 304, 309–310

and Euler characteristic, 372and Gauss map, 309–310

Total geodesic curvature, 364–366Total rotation, 380Transferred frame field, 290–291Translation of Euclidean space, 100–101, 113Trefoil knot, 80–81 (Ex. 19), 235 (Ex. 21)Triangle, 378–379, 441 (Ex. 4)Triangle inequality, 286 (Ex. 3)Triangulation, 370Triple scalar product, 4850Tube, 234 (Ex. 17), 235 (Ex. 21)

2-segment, 176–177

U

Umbilic point, 212–213, 233 (Exs. 14, 15).See also All-umbilic surface

Unit normal function, 225–226Unit normal vector field, 186, 203Unit points, 36Unit-speed curve, 53Unit sphere, 131–132Unit tangent, 58, 69, 72–73Unit vector, 45

V

Vector, See Tangent vectorVector analysis, 33 (Ex. 8)Vector field

on an abstract surface, 195on a curve, 54, 332–333on Euclidean space, 8on a surface in R3, 153,in terms of a patch, 158 (Ex. 12)

502 Index

normal, 153–154tangent, 153

Vector part, 6Velocity (vector), 18, 195Vertices, 366Volume element, 33 (Ex. 6)

W

Wedge product, 29–30, 160Winding line on torus, 158 (Ex. 11)Winding number, 181 (Exs. 5, 6), 191–192,

385–386

Index 503

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Elementary Differential Geometry · 2019-03-02 · Elementary Differential Geometry R evised Second Edition Barrett OÕNeill Department of Mathematics University of California, Los